
Class. 
Book. 



IghtN 



COPYRIGHT DEPOSIT. 



MARYLAND INSTITUTE 
HANDBOOK 



Written and Compiled for the Use 
of the Students by 

A. O. BABENDREIER 




Published by 

The Maryland Institute for the Promotion of 

Mechanic Arts and Design, 

Baltimore, Md. 



c& 






Dedicated to 

HENRY ADAMS, M. E., 

Chairman of Committee, Market Place School, 

in appreciation of his untiring efforts 

in promoting the interests of 

the School 



Copyright 1920 

The Maryland Institute for the Promotion of Mechanic Arts, 

Baltimore, Md. 



©CLA566069 



■ i 
i 



U 

/Hi 



PREFACE 



The Maryland Institute Schools of Art and Design, of 
Baltimore, Maryland (NIGHT SCHOOLS) provides a 
course in Mechanical Drawing, the students of which 
vary in age from 16 to 50 years, and are recruited from 
all classes of workers: shopmen, clerks, apprentices, and 
young men still attending day schools. These find that in 
their work a knowledge of mechanical drawing and some 
knowledge of mechanics are essential. Many, due to lack 
of application, have forgotten much of that learned at 
school, while the majority have attended only the lower 
grades of the common schools. 

The object of the following pages is, in so far as possi- 
ble, to help these students. The course, however, being 
primarily for teaching mechanical drawing and rela- 
tively short, the time allowable for mechanics is very 
limited. In view of this, only the very elementary prin- 
ciples are considered, and then only to a point from which, 
coupled with the lectures, problems, etc., the student may 
further aid himself by use of more advanced text and 
handbooks, afterward referred to. 

In the matter of the strength of materials, the meaning 
of the various technical terms generally used and their 
relationship one to another are explained; also the appli- 
cation of simple formulas are shown. These with the 
lectures, examples and problems by the instructor, should 
so fit the student that he may open and use a handbook 
without the usual fear and anxiety experienced upon 
entering an unknown field. 



It is earnestly hoped the student will follow up the 
work by further study, for nothing can be gained without 
effort. (So keep working.) 

The more knowledge you have of a subject, the more 
interesting it becomes. (So seek.) 

What man has done, man can do again. (Do also.) 

As man is imperfect, so, too, are machines. (Improve 
yourself and try to improve that with which you have to 
deal.) 



CHAPTER I 



DEFINITIONS 



MECHANICS 

This word is used in two ways: 

One to indicate that class of skilled workers, machinists, 
carpenters, bricklayers, etc., who use tools and imple- 
ments to produce new machines or objects. 

The other and broader meaning — the "science of forces 
and powers, their action and application. " 

MACHINE 

This term is also used in two ways: 

One to indicate all tools from the simplest to the most 
intricate machine. 

The other to all apparatus used to modify or regulate 
the effects of natural forces. In either case the machine 
may be simple or complex, depending upon the work it 
has to do, whether it be to modify a force or modify a 
motion. 

FORCE 

Any cause which produces or tends to produce motion 
or a change of motion. 

Force is invisible and is known only by the effect 
produced. 

GRAVITY 

One force with which man has continually to deal. 
This natural force — like all other natural forces — acts 
always in the same way under the same circumstances. 



6 Elementary Mechanics 

It is the attraction between all objects and the earth. 
Its effect is shown by causing a body to have weight. It 
acts always in a straight line and toward the center of 
the earth. 

An object held in the hand, i. e., a glass, book, or piece 
of iron, is said to have weight, and must be supported to 
counteract the force of gravity to prevent it from falling 
to the ground, and in a straight line at that. 

CENTER OF GRAVITY 

A point within the object from which its entire weight 
may be suspended or supported by a single vertical force. 

Whenever any object is suspended by a single cord 
attached thereto, the center of gravity will lie somewhere 
in the body in line with the cord. 

Forces other than gravity are met with in mechanics, 
either singly or in groups and in direction other than 
toward the center of the earth. 

ACTION AND REACTION 

To every action there is a reaction of like intensity 
and in the opposite direction. It takes as much force 
pushing upward to sustain the glass, book, or iron as they 
push downward when held in the hand. If these two 
forces were unequal, the object would move in the direc- 
tion of the greater one. If the object be suspended by 
a cord, the reaction would be through the cord, and for 
its entire length. 

Forces are also known as Internal or External — the 
former acting in the object tending to change its form, 
and the latter acting from the outside, either tending to 
change its position called "Pressure" or actually moving 
the object called "Moving Force." 



Elementary Mechanics 7 

Internal forces may be active or reactive — active when 
the particles of matter tend to separate, as in steam and 
gases; reactive when resisting destruction against forces 
externally applied. 

Magnetism acts between the particles of matter form- 
ing the body, and therefore is classed as an internal force. 

EQUILIBRIUM 

When two forces act one against the other and no 
motion is produced, the forces are in equilibrium or 
balanced. 

Forces being invisible, we know them only by the 
effects they produce, and then only when we can measure 
the intensity and know the direction in which they act 
and the point on which they act. The intensity is meas- 
ured by a weight or its equivalent. The direction is 
opposite to that in which the counteracting force acts. 

In this country the unit of measurement is the pound 
or the ton (2,000 lbs.) — depending whether the force be 
small or large. 

RESULTANT 

Two forces or more acting on a single point may be 
combined, so that one counteracting force may hold them 
in check. Such combined force is called the Resultant 
force. 

If the two or more forces act in the same direction and 
in the same line, the resultant is equal to the sum of the 
forces and equal to the counteracting force acting in the 
opposite direction. 

If two forces of different intensity act on the same point 
but in opposite directions, the resultant is equal to the 
difference between the two forces, and acts in the same 



8 Elementary Mechanics 

direction as the greater one and is equal to the counter- 
balancing force acting in the opposite direction. 

When two forces act upon the same point other than 
in a straight line, i. e., at an angle to each other, the 
resultant is neither equal to the sum nor the difference 
between the forces, nor in the direction of either, but a 
combination of both. 

Two methods are used to find the resultant, either by 
trigonometrical formula or by graphic representation. 

The latter is the more convenient, and is frequently 
referred to as the parallelogram of forces and force 
diagram. 

FORGE DIAGRAM 

A force can be measured, and a line may represent it 
by its length and direction, if a unit of length (1 M ) rep- 
resents a unit of force (lbs.). 

Take two forces A (200 
lbs.) and B (300 lbs.) acting 
on the same point with an 
angle between them. Let 1 ,! 
represent 200 lbs. of force, 
an arrow point the direction 
and "C" point of application of the forces. Then AC — 
200 lbs. or 1 inch, BC = 300 lbs. or lj inches. From C 
draw BC= lj inches parallel to force B and draw AC = 1 
inch parallel to force A. 

From A and B draw AD parallel to BC and BD 
parallel to AC, and from C draw the diagonal CD to the 
point of intersection; then CD is the resultant in direc- 
tion and amount, and its length in inches multiplied 
by 200 lbs. (unit of length) gives the resultant force in 
intensity and direction. 




Zoo-UV 




Elementary Mechanics 9 

If more than two forces in the same plane act on one 
point, proceed as before, taking two of the forces and 
finding the resultant; then combine this resultant with 
the next force and again find the resultant; continue in 
this manner until all the forces have been taken. The 
last resultant found is the resultant of all the forces. 

An easier method is to arrange the forces (see Fig. 2) 
about the point of application, and proceed as follows: 
From the end of force No. 1 
draw 2a y parallel to 2; from 
end of 2a draw 3a, parallel to 
J; then from end of 3a draw 
resultant R to point of appli- 
cation 0. To hold the forces 
in equilibrium, a force applied 
at in opposite direction to the resultant R and of equal 
intensity is necessary. If more than three forces, use the 
same method until all have been considered. 

INERTIA 

The inability of a body to change its state of rest or 
motion. 

WORK 

The result of a force acting upon a body and producing 
motion is called work. The unit of work is the foot- 
pound, or the force of one pound moving a body through 
one foot of space. 

Work (ft.-lbs.) = lbs. weight X ft. lifted. 

Example. Thus 1 lb. weight lifted 10 ft. = 10 ft.-lbs. work 
Also 10 lbs. weight lifted 1 ft. = 10 ft.-lbs. work 
or 5 lbs. weight lifted 2 ft. = 10 ft.-lbs. work 



10 Elementary Mechanics 

For the purpose of comparing two operations, the work 
done may have been the same, yet the time for doing 
them may have been different. 

Example. 100 ft.-lbs. work done in 1 minute as against 
100 ft-lbs. work done in 5 minutes 

Here the work done was equal, but the one took five 
times as long as the other. It will be seen that the rate 
of doing the work must be considered. This leads to the 
term power, which takes the time into account. 

POWER 

The product of the force (lbs.) multiplied by the dis-?' 
tance (feet) multiplied by time (minutes consumed equals 
the power. 

Expressed. P = Force X Distance X Time. 

The unit of power used is the horse power (H. P.). 
This unit represents 33,000 foot-pounds of work done in 
one minute of time, and was adopted by James Watt for 
comparing the work of his steam engines with the work 
that could be done by London draft horses. 

Example. 1000 lbs. raised 33 ft. in 1 minute, or 33,000 ft.- 
lbs. per minute equals one horse power. 

MECHANICAL POWERS 

Are those elementary machines used by man to modify 
motions and change the magnitude of forces, as follows: 

Lever — This is the most commonly used of the me- 
chanical powers, and consists of a bar supported at some 
point called the fulcrum. Its use is to modify a motion 
or force by means of power applied; the power and 
weight respectively being applied at points distant from 




p 




WF- 



Elementary Mechanics 11 

the fulcrum. These distances are known as the power and 
weight arms, respectively, and must be measured at 
right angles to the direction in which the power and 
weight, respectively, act and are not necessarily the 
actual physical lengths of the arms. The wheel and axle 
and gear wheels embody the principle of the lever also. 

The principle is applied in three ways, known as classes 
(see Fig. 3): 

Class 1. In which 
the fulcrum is be- 
tween the weight 
and the power, 
which move in \^_ 
oppositedirections. rims 2 

Class2. In which 

the weight is ap- « pf 

plied between the 
power and the ful- 
crum. Here both 
weight and power 
move in the same 
direction. 

Class 3. In which the power is applied between the 
weight and the fulcrum. Here also the weight and power 
move in the same direction. 

In Class 1 the power and weight may be equal or 
unequal, and either the one or the other may be the 
greater, depending upon the relative lengths of the two 
arms. 

In Class 2 the power may be equal to or less than the 
weight, depending upon the lengths of the arms. 

In Class 3 the power must equal or be greater than the 
weight. 




FIG. 3. 



12 Elementary Mechanics 

The actual weight of the lever itself and friction are 
not considered in the study of the action of levers. , 

Law — One law governs all three classes and may be 
stated as follows: 

Weight (W) multiplied by the Weight Arm (JVF) 
equals the Power (P) multiplied by the Power Arm 
(PF). Where F represents the fulcrum. The formula 
is generally expressed, 

WXWF = PXPF 

If any three of the above are known, the fourth may be 
readily found. 

Examples. Class 1. /F = 50lbs. WF = 4tit. PP = 8ft. 

Then 50X4 = PX8 or 

P = — - or 25 lbs. 

o 

Class 2. /F=50lbs. /FP = 4 ft. PF = S ft. 
Then 50X4 = PX8 or P = ~=25 lbs. 

o 

Class 3. /F = 50lbs. /FP = 8ft. PP = 4ft. 
Then 50X8 = PX4 or P = ^9=100 lbs. 

Levers may be straight, bent or curved. The arms in 
any case must be considered as before stated, i. e., the 
straight distances from the fulcrum to points opposite 
the weight and power, respectively, and at right angles to 
their lines of direction. (See broken lines in Fig. 4.) 

Pulley — The pulley is used where bodies (weights) 
are to be moved over longer distances, and consists of a 
wheel over which a rope or its equivalent is passed; to 
the one end of which the weight is attached and to the 
other end the power is applied. 



Elementary Mechanics 



13 




Pulleys are arranged 
singly or in pairs or 
groups. The center or 
pin on which a pulley 
turns is called the Axis. 
The pulley may be known 
either as Fixed; if its axis 
does not move with the 
weight; or as Movable, 
if its axis does move with 
the- weight. 

A fixed pulley changes 
the .direction of the pull 
only, and has no me- 
chanical advantage, be- 
cause the weight and 
power move over the same distance. 

A movable pulley has a mechanical advantage, because 
the weight moves only one-half as fast as the power or 
one-half the distance over which the power moves. 

The fixed pulley is an example of a lever of the first 
class, with the weight and power arms equal. 

The movable pulley is an example of levers of Class 2 
or 3, i. e.: 

If of Class 2, one end of the 
diameter of the pulley represents 
the fulcrum; the axis the point of 
application of the weight, and the 
opposite end of the diameter the 
point of application of the power. 
The power arm is, therefore, twice 
as long as the weight arm and the 
power will consequently move twice 




14 



Elementary Mechanics 







as far as the weight with one-half the effort (see Fig. 5). 
The same amount of work is done whether the weight is 
lifted direct or whether a fixed or movable pulley is used; 
as weight X distance lifted = power X distance through 
which it acts. 

The advantage derived from a fixed 

pulley is the change from a lift to a pull; 
the advantage from the movable pulley 
is that only one-half the effort but ex- 
J erted over a longer distance is required. 
The arrangement shown in Fig. 5A has 
no advantage over the single movable 
FI6SA ' pulley, and the fixed pulley is provided 
y? w merely to change the direction of the pull. 

Block and Tackle — A combination of ropes 
and pulleys (sheaves) grouped together, usually 
consisting of one or more pulleys in a fixed 
block and one or more pulleys in a movable 
block, the latter attached to the weight to be 
lifted and a rope, one end of which is attached 
to the fixed block and then passed over mov- 
able and fixed pulleys alternately until all 
have been served; the free end acts as the 
pull rope. The number of passes of the rope 
moving toward the fixed pulleys and away 
from the movable pulleys produce a mechan- 
ical advantage, as will be understood from 
Class 2 of levers. (See Fig. 6a.) 

The advantage to be derived may be found 
from the formula in which P represents 
Power; /F, Weight; and R, the No. of Ropes 
moving away from the movable block and 
toward the fixed block. 



Elementary Mechanics 



15 



Then Power = Weight divided by twice the number 
of ropes moving away from the movable 
block and 

w 



p= 



2XR 



Example. A weight of 240 lbs. is to be lifted with a group 
of pulleys consisting of 3 fixed and 3 mov- 
able. What power is required on the pull- 
rope, and how many times the distance the 
weight moves must the end of the pull-rope 
traverse? 



Solution. 



P = 



240 
2X3 



= 40 lbs. and pull-rope moves through 



six times the distance the weight moves. '** 



The formula for any arrangement of 
pulleys is often given in the form 

Power = Weight divided by the total 
number of ropes, less one. (It is to be 
understood that here, the last pass is sup- 
posed to have passed over a fixed pulley 
last.) 

In the same example as before: 



W 



or 



240 240 



No. Ropes-1 7-1 



= 40 lbs. 



In Fig. 6b. The first ropes (x) (xi) and 
the arrangement do not satisfy Class 2, but 
do satisfy Class 1 of levers, and therefore I 
the extra pulley "0" is required to produce v 
the same direction of pull as in Fig. 6a. 
The end of the rope, therefore, would be 



16 



Elementary Me< bank b 



better placed if attached to the fixed block and not to 
the movable block as shown. 

Note. — If the ropes passing over a movable pulley are not parallel 
and diverge from the pulley, some of the mechanical advantage is 
lost. When said divergence reaches 120 degrees — all the mechanical 
advantage is lost. The power (P) at any one instant may be found 
in two ways: 

One by the formula: Power equals one-half the weight 
multiplied by the secant of the angle "a" or 
P = \W (Sec. "*") (See Fig. 7.) 



IVERGC* C< 




FIG 7. 



IW 




w FiG 7\ 







The other by the use of the force diagram (see Fig. 7A), 
wherein the two cords leaving the movable pulley are 
represented by two of the sides of the parallelogram with 
which they are respectively parallel and the weight is 
represented by the length "tf." This 
shows one of the main' ways in which 
the force diagram may be used to solve 
various problems. 

DIFFERENTIAL PULLEY BLOCKS 

This machine (sec Fig. 8) is used for 
lifting heavy weights. A chain must be 
substituted for the rope to avoid slip and 
in the fixed block two chain wheels of 
different radii R and r, one larger than 
r,6 «i the other, are used; both are on the same 




Elementary Mechanics 



17 



spindle and fixed together so that one cannot turn with- 
out the other. The chain is continuous, starting from 
the movable pulley; thence over large fixed wheel to pull- 
ing section; thence (slack chain) to the small fixed wheel 
and back to the movable pulley. 

The power X large radius R = one-half the weight 
multiplied by the difference between the large radius, 
R, and small radius, r, of the two fixed pulleys. 

W(R-r) ' 



PXR = W(R~r) or P = 



2R 



Example. /F = 240 lbs. R = 9" r = 6" 



Then PX9 = 



Z lo 



WINDLASS 



■ — „ i. 




F\G.9a 



A machine embodying the principle of 

levers under the term of wheel and axle. 

It has been and still is extensively used 

in place of pumps on farms to lift buckets 

of water from wells, also by the builder 

to lift his materials of construction. It 

consists of a drum or cylinder fixed on 

an axle and made to rotate by a pulley, 

crank arm or a series of lever arms also 

fixed on the axle. As the drum is made to turn, 
the lifting rope with one end fixed to the drum 
and the other end to the weight to be lifted, is 
gradually wound upon the drum, thus causing 
the weight to be lifted. The radius of the 
drum constitutes the weight arm, the axle pro- 
vides the fulcrum and the radius of pulley or 
the length of the crank arm provides the 
FI6.9V power arm. If a gear (toothed wheel) is used 




18 Elementary Mechanics 

in place of the pulley or crank, each tooth would repre- 
sent a lever arm which would be successively brought 
into action. (See Fig. 9a-b.) 

The same law that governs levers also applies here — 
Power X Power Arm = Weight X Weight Arm 
ovPXR = WXr 

Also P™. and W= P -** 
K r 

Example. Weight = 100 lbs. R = 15 M r = 3 

_ 100X3 _,, 

Then P = — — — =20 lbs. 

Now as the weight in pounds multiplied by the dis- 
tance in feet it is raised = the ft.-lbs. of work performed; 
so also must the power in pounds multiplied by the dis- 
tance in feet it acts = said work, and therefore the power 
and weight will be in the same proportion as the relative 
distances traversed. This holds equally true in the cases 
of levers, pulleys and wheel and axle; the mere fact that 
in the latter we have the circumferences of the drum and 
wheel to consider does not alter the relationship, as may 
be shown by the example as above: Weight 100 lbs., 
Radius of wheel 15" (R) or 30" diameter, Radius of drum 
3" (r) or 6" diameter. 

Now for one revolution we have: 

Work performed in ft-lbs. = 100 lbs.X6 M -M2 
X3.1416 
Also Work performed in ft.-lbs. = Power(lbs.) X30" 
-M2X3.1416 
100X6X3.1416 PX30X3.1416 . 



And 



12 12 

D 100X6X3. 1416-f- 12 ,, 

P= = 20 lbs. 

30X3.1416^-12 



Elementary Mechanics 19 

Ratio of circumference of drum to circumference of 
wheel, as will be seen, is as 1 to 5 and power equals one- 
fifth the weight, or 100^5=20 lbs. 

INCLINED PLANE 

This one of the mechanical powers consists of an 
inclined surface to facilitate the lifting of an object or a 
weight from a low surface to one higher where it would 
be inconvenient or impossible to lift the weight ver- 
tically — as for example, the transfer of heavy boxes and 
barrels from a pavement to the platform of a wagon; the 
movement (lifting) of a car from the bottom to the top 
of a hill; the splitting of wood by an ax — the sides of the 
ax (wedge) are the surfaces of the inclined plane. 

In the study of the inclined plane two methods may be 
used for finding the power necessary to hold in position 
a weight thereon, and also the intensity with which the 
weight tends to break through the surface. It will be 
seen readily that if the plane is nearly vertical the ten- 
dency to break through it would be small, but the lift 
would be more direct and, too, the plane would be short 
but steep, and the power would nearly equal the weight. 
On the other hand, if, with the same height the plane 
be long and therefore nearer horizontal, the reverse of 
the above would be true. 

The two methods for finding power and pressure are: 
one the application of the parallelogram of forces and the 
other by the application of trigonometry, or the relation 
of the sides of the triangle, i. e., the length of inclined sur- 
face, the vertical height and the horizontal distance. 

If, now, we take the diagram (Fig. 10), A-B is the in- 
cline; A-C, the horizontal distance; B-C, the height; and 
W, the weight, concentrated at its central point or cen- 




20 Elementary Mechanics 

ter of gravity. The weight W in lbs. may be repre- 
sented by the length of a vertical line W-E as the 
resultant of the parallelogram; the two sides of such 

3 parallelogram Wx and Wy 
may be found readily, the 
former being parallel to the 
plane and the latter at right 
angles thereto. Now, by 
taking the same unit of 
length per lb. or per 100 lbs., which indicated the weight; 
Wx and Wy will represent the intensity with which the 
weight tends to slide down the incline and the tendency 
to break through the plane at right angles to the surface. 
The intensities represented by Wx and Wy would equal 
the power required in each case but acting in the 
opposite direction. 

The Second Method — The power P required to pre- 
vent the weight W from sliding down the incline is 
weight multiplied by height in feet, B-C, divided by the 
length in feet, A-B) of the incline, or 

p WXc . p WXd 
r = — : — ' also r\= — - 7 — 

O 

where P = the tendency to slide down the incline in 
pounds 
Pi = the tendency to break through 
W = Weight in pounds 
b = the length A-B 
c = the height B-C 
d = the length A-C 
c and d, respectively, represent the sine and cosine of 
the angle "a"; also sides b, c and d bear the same relation- 



Elementary Mechanics 21 

ship to each other as the sides and diagonal WE, Wx and 
Wy of the parallelogram of forces. 

Note. — The force P as found would in practice be more than neces- 
sary to prevent ihe weight sliding down the incline, but less than 
actually required to move the weight up the incline, because, first, we 
have merely provided for balancing the force exerted down the in- 
cline, and, second, we must provide also sufficient extra power to 
overcome the friction between the weight and the inclined plane, 
which retards movement of one object upon another and differs with 
the kinds of material in contact, with the nature of the surfaces in 
contact and the pressure on the surfaces. 

FRICTION 

The resistance to the movement of one body upon 
another. As a force is required to overcome it, it is often 
referred to as a force. The force required to overcome 
friction varies with the pressure, the nature of the surfaces 
in contact and the materials. 

Sliding Friction — If two blocks of wood, each of same 
size and weight, one rough and the other smooth, be 
placed and made to slide on a rough board, a greater 
effort is required to keep the rough block moving than is 
required for the smooth one. If the smooth block be 
made to slide on the same board, but the surface thereof 
be made smooth, even less effort is required. This could 
be further reduced if a lubricant be introduced between 
the surfaces. Friction is also less if the materials are of 
different natures, also if the surfaces be hard and un- 
yielding. 

As the surfaces and nature of the materials in any one 
case may vary somewhat, the actual force required may 
vary also. In order to compute the force to overcome 
friction certain co-efficients are used, represented by the 
letter "/." These coefficients are given as a percent, or so 
many 1-100 parts (expressed by decimals) of the pressure 



22 Elementary Mechanics 

between the surfaces for the varying conditions met with 
in practice. These coefficients "/" were deduced through 
experiment by using a plane surface or board, placing 
the weight thereon and then raising one end of the plane 
until the weight begins to slide; the angle of inclination 
of the plane to the horizontal is called the angle of repose, 
designated by 'V (see Fig. 10 also). 

The tangent of this angle equals the coefficient of 
friction "/." 

This is expressed by the formula: 

. c . c Ey Wx 
f =d' hut d = Wy 0I W^ 

but Wx represents the force along the plane, and Wy the 
weight normal to the plane. The actual weight of test 
block is represented by the line WE, and in compliance 
with the laws of the inclined plane, the weight normal to 
the plane is Wy and equal to the cosine of angle "#'» 
times the weight, and the force Wx along the plane in 
like manner is equal to sine "a" times the weight. 

Now, if the normal pressure Wn of a body sliding on 
a surface multiplied by the coefficient of friction for any 
condition of material and surface is found, this will 
represent the power P{ required to overcome the friction 
alone. It may be expressed by the formula: 

A few coefficients (/) taken from "Machinery Hand- 
book" give: 

Bronze on bronze 0.20 

Bronze on cast iron 0.21 

Cast iron slightly lubricated. ... 0.15 

Hardwood on hardwood dry. . . . 0.48 

Leather on hardwood dry .33 

Leather on cast iron 0.56 



Elementary Mechanics 



23 



Other coefficients may readily be obtained from any of the 
recognized handbooks. 

Note. — The above conditions hold good whether the surfaces be 
flat and slide one upon the other or whether the surfaces are curved 
as in journals. The friction of rest is somewhat greater than the 
friction of motion — which is readily perceived in starting a sliding 
piece from a state of rest. 

The laws of friction: 

1. Under moderate pressures the friction is propor- 
tional to the pressure — with excessive pressures the co- 
efficient of friction rises rapidly. 

2. The total friction is independent of the area of the 
surface under moderate pressures. 

3. Under low velocities the friction is not dependent 
upon the velocity, while under high velocities the friction 
decreases. 

ROLLING FRICTION 

When a wheel rolls along a surface, the perimeter of 
the wheel and surface are each somewhat depressed at 
the point of contact, which necessitates what may be 
termed a constant climb- 
ing due to this depression 
of the surfaces. It will 
be readily seen (see Fig. 
1 1 A) that the greater the 
diameter of the wheel the 
less will be the force re- 
quired for the climbing 
action due to the increased length of the lever arm, as 
represented by the radius of each of the two circles repre- 
senting two wheels with the power applied at the axle in 
each case. In like manner, the same holds true when the 
power is applied at the top of the wheel (see Fig. 11B). 





FIG. 1 1 A. 



FIG. M B. 



24 



Elementary Mechanics 



When the power is applied to the axle, sliding friction 
must also be considered, as this forms a bearing. Where 
the power is applied to the rim, rolling friction for the 
upper surface must be taken into account. 

THE SCREW 

This machine in practice really combines two of the 
mechanical powers, viz: the inclined plane and the lever, 
as in the Screw Jack, Fig. 12c. If an inclined plane (see 
Fig. 12a) be wrapped around a cylinder, the base of the 




F's-igg 




FIG. 12 b. 



plane — equal to the circumference of the cylinder and 
the height of the plane equal to the forward movement 
of the screw in one revolution, or lead, its upper edge 
will form the helix (see Fig. 12b). In practice the 
inclined plane is arranged as a screw thread, in that the 
screw may accommodate a nut with similar threads on 
the interior. 

If a weight is supported by the screw and lifted 
thereby, during one revolution the screw will advance 
the distance represented by the lead and will lift the 
weight through a like distance. Now, if the lead were 
1 inch for each revolution, and the weight were 2,400 lbs. 



Elementary Mechanics 



25 



— the work done would equal the weight lifted in pounds 
times the distance moved in feet, or: 

P = 2,400 lbs.Xl"-M2 = 200 ft.-lbs. of work 
Power in ft.-lbs. (P) = weight (W) X distance in 
feet (D) neglecting friction. 

Now, as a wrench or lever L (Fig. 
12c) is necessary to turn either the 
screw or its engaging nut, the length 
of such wrench or lever in feet mul- 
tiplied X 2 X 3.1416 would give 
the distance the lever arm has 
moved to do the 200 ft.-lbs. of work 
in one revolution. Now, if the lever 
were 2 feet long, the distance trav- 
ersed would equal 2 X 2 X 3.1416 = 12.5664 ft. and the 
power (Pl) required at the end of the lever would be: 

P 




Pl = 
Pl = 



ZX2X3.1416 
200 



or 



12.5664 



= 16 lbs. about 



But on account of the friction between the threads the 
power (Pl) would have to be increased. Also the friction 
either at the end of the screw or the bottom surface of 
the nut would have to be considered, depending upon 
which one is turned. 

For purposes of construction and strength, screws may 
have one or more threads and be known as single, double 
or treble, etc., threaded. In any case, the distance in feet 
traversed by the screw in one revolution must be consid- 
ered and this is determined by any one thread. 

In the problem given, the weight for one revolution of 
the screw was lifted 1 inch or 2,400 inch pounds, but as 



26 Elementary Mechanics 

work and power are usually in foot-pounds, the inch- 
pounds (2,400) must be divided by 12 (inches per foot) 
to obtain foot-pounds of work done. If inch-pounds are 
considered, the length of the lever arm L in the second 
part of the problem should be expressed in inches in 
place of feet, i. e., 24 inches. 

By combining the two parts of the formula 

WXD 2400X1" = 50X1 . 16 lbs. 

L LX2X3.1416" 24X2X3.1416 "3.1416 nearly 

D and L being expressed in the same unit of length. 

If in any case, the power at the end of a lever; the 

length of the lever and the lead of the screw are known; 

the weight which may be lifted, neglecting friction, may 

be found by rearranging the formula and using it in the 

form: 

/// = PXZ>X2X3.1416 

D 

Example 1 — What weight can be lifted by a screw if a 
pull of 8 lbs. (P) is exerted at the end of a lever arm 24" 
long (L) which turns a single thread screw having 2 
threads per inch or a lead of J" (D) per revolution? 

r = 8X24X2X3 1 1416 = 8x24><2><3 J416X2 = 2412 74 

2 

Note. — The result would have been the same had the power been 
16 lbs. and the lever arm 12 inches, or had the power been 16 lbs. 
and the lead 1" (1 thread per inch). 

MOTION 

A body moving or changing its relation to another 
body is said to be in motion. A ball thrown from the 
hand, an automobile running along the road, or a wheel 
turning on its axis, are each in motion relative to some 



Elementary Mechanics 27 

other object; such as the earth or objects thereon or as 
the wheel relative to the axle. 

No body will change its position and motion result 
unless acted upon by some force. A force always tends 
to produce motion or to change the direction thereof. 

Laws of Motion — To Sir Isaac Newton, an English 
scientist, we are indebted for three laws, as follows: 

First — Every body continues at rest or in uniform 
motion in a straight line unless acted upon by some force 
which may tend to change said state of rest or uniform 
motion, 'V 

Second — A change in motion and the direction of said 
change is proportional to and in the direction of the 
force producing the change "b". 

Third — To every action there must be an equal re- 
action and from the opposite direction "c." 

'V — The ball, the automobile or the wheel has no 
power to start moving, or if moving, to move faster, or 
slower, or to stop unless acted upon by some force. 
Weight (gravity) and a push or pull tend to produce 
motion or increase it. Friction and work tend to retard 
or to destroy it. 

Increasing the motion is called "Acceleration." De- 
creasing the motion is called "Retardation. " 

"b" — A moving object, if acted upon by a force at an 
angle to the direction of its movement, will be deflected 
from its course in proportion to the intensity of the 
second force. This can readily be seen in the resolution 
of forces. 

"c" — A ball held in the hand is pushed upward with 
the same force as its weight, due to gravity, tends to push 
the hand down. 



28 Elementary Mechanics 

The fulcrum of a lever pushes in the opposite direction 
and with the same intensity as the forces which tend to 
move it, If this were not true, in each case movement of 
the fulcrum would result due to unbalanced conditions. 

As time entered into consideration with "work" and 
"power (HP)" so also must it be taken into considera- 
tion when dealing with motion — that is, the amount of 
motion or distance covered in a unit of time or its rate 
must be considered. This rate of motion is called 

velocity. 

VELOCITY 

If two trains both alike go in the same direction and 
for the same distance on parallel tracks with no stops, 
and the one takes less time than the other, its rate of 
motion or its velocity would consequently be greater 
than the slower train. 

In considering the speed or velocity of trains, the dis- 
tance is usually expressed in miles and the time in hours; 
and the rate of speed in "miles per hour." 

In steam engines, it is the travel of the piston in "feet 
per minute." 

In mechanics, velocity is expressed in "feet per second" 
for the reason that in this form we can compare it with 
the velocity resulting from a like object falling verti- 
cally due to the force of gravity. 

If a train runs 60 miles in 2 hours, its rate would be: 

^(miles per hours) = — - or 30 miles per hour. 

The rate in feet per second is usually expressed by the 
plain letter "F." 

Now, in one hour there are 3600 seconds, and in 30 
miles there are 30X5,280 ft. or 158400 ft. 

Then 3600 F= 158400, and F= 15 ^5? - or 44ft. per sec. 

3600 



Elementary Mechanics 29 

Again, if an object moves over a distance of "A" feet in 
"t" seconds, whether circling around a central point or 
whether in a straight line, the velocity in feet per second 
may be expressed: 

TTt 1 J! J\ f eet W 

/Mvel. it. per second) = ~ t-t- 

bec. {t) 

h 
or y =- 

Example — A body travels 100 ft. in 5 seconds; there- 

, „ 100 „„ , , 

fore V = — =— or 20 feet per second, 
o 

The same formula may be expressed t = yy or h = t X V, 

depending upon which factor is required. It is under- 
stood two of the factors must be known. 

Take the above example. If the velocity and distance 

passed over had been known, the time t = — - or 5 seconds. 

In like manner, if the velocity and time had been given 
to find the distance gone over, then the distance h = 5 X20 
or 100 ft. 

In the above examples, both the train and the object, 
when moving at the velocities as found, i. e., 44 ft. and 
20 ft. per second, relatively, would, according to Newton's 
first law of motion, continue at this rate forever, pro- 
vided no external force act thereon either to increase 
(accelerate) or decrease (retard) their motion. Nor 
should any power be required to keep up this rate. The 
above is not obtainable, as the resistance of the air, the 
sliding and rolling frictions of the journals, and track, 
all help to retard the movement of the train. In a like 
manner, the friction between the parts of a steam 
engine, line shafting, machine and the final work to be 



30 Elementary Mechanics 

done, all call for the expenditure of energy; so that, to 
keep the train or the engine moving at the uniform speed, 
it requires the constant application of sufficient power 
to overcome such resistance or load, otherwise the train 
or engine would slow down and finally stop. On the 
other hand, if part of the load were relieved without an 
equal reduction in the power applied, the speed or 
velocity would increase or accelerate until the parts, 
especially the flywheel of a stationary engine, would fly 
apart with disastrous results. To provide against the 
slowing down on the the one hand and against excessive 
speed on the other, engine governors are used. The 
governor acts also to conserve the energy stored in the 
steam. If the excess of power above that required 
remain uniform, the velocity would increase at a uniform 
rate. 

VELOCITY OF FALLING BODIES 

Gravity, as before stated, is the force of attraction 
between all bodies and the earth, and causes them to 
have weight. This force, too, naturally causes an accel- 
eration in the speed or velocity of the falling body, and 
at a uniform rate of 32.16 feet for each second of time. 
If a body starts falling from a state of rest, its velocity 
at the start would be feet per second; at the end of the 
1st second its velocity would be 32 feet per second (in 
round numbers); at the end of the 2nd second it would 
be 64 feet per second; at the end of the third, 96 feet per 
second, and so on. 

If the velocity at the beginning of the 1st second is 
ft., and at the end of the 1st second is 32 feet, the space 
passed through during 1st second is 16 ft. or (0-jr32) 4-2 = 
16 ft. per second — this being the mean velocity for the 
1st second. 



Elementary Mechanics 31 

The mean velocity for the 2nd second is (32 + 64) 4-2 = 
48 ft. per second; and mean velocity for 3rd second is 
(64+96) -s-2 = 80 ft, per second, and so on. Each second 
the velocity increases 32.16 feet per second. 

It will also be seen that at the end of any number of 
seconds of time the velocity would equal the number of 
seconds multiplied by the rate of acceleration, 32 feet 
(32. 16' actually) per second — so if V represents the veloc- 
ity in feet per second; t, the time; g, the effect of gravity, 
then V — gt, and the velocity at the end of the 3rd second 
would be 

F=3X32 or 96 ft. and the mean velocity designated 

1 2 96 
V m for the full period would equal V m = -^ or — =48 ft. 

and if h represents height of fall in feet, then 

h= V m Xt or for 3 seconds A = 48X3 or 144 ft. 

t g 
but V m (as before) = — , so that the formula may be 

expressed 

h=^X torpor \gi 2 

V 

As V = gt and t — — , by substitution in the preceding 

o 

equation 



(El\ z 11 11 

g V e 2 ) 1 ? 2 e V 2 

h= -Af^ = ±J- = 4 and h= ~ and 



2 2 2 2g 

2gh=F 2 or V= V 2gh 
Now, as g = 32.16, then 2g = 64.32 F= V64.32XA, but 
the square root of 64.32 is 8.02, so the formula is often 
used in the form 

T=8.02Vr 



32 Elementary Mechanics 

Now, if the distance "A" through which a body is falling 
or moving is known, its velocity by the last formula may 
readily be found — as, for instance, it was ascertained that 
in three seconds of time the distance A was found to be 
48X3 or 144 ft. 

48 ft. being the mean velocity and 3 the time in seconds, 

F = S.02^Jh or V= 8.02 Vl44 or 8.02X12 or 96.24 ft. 

96.24 
per second, or a mean velocity ( F m ) of — '- — = 48. 12 ft. 

Should the time 'V be required, the above formulas 
may be written 

^_ \2h a/A -i 



t = 



l = \2h_ -s 



g 4.01 

From the above it will be seen that a moving body, 
whose weight and velocity are known, may be compared 
to a falling body of the same weight falling through a dis- 
tance of A feet, and that it can do the same amount of 
work as such body falling the same height. 

All bodies would fall equally fast in a vacuum, but fall- 
ing in the air their movement is retarded — those of 
greater density falling faster, due to their volume 
being relatively smaller. A lead ball and a feather will 
fall equally fast in a vacuum, but in air the lead ball will 
fall much faster, as the air would offer less resistance to 
its passage. 

ENERGY 



aw> 



If a body of any weight"/F" is lifted through "A' 
feet, the amount of work in ft.-lbs. expended would be 
E = Wy,h. If the weight were allowed to fall, it would 
do an amount of work equal to that required when 
raising it; therefore the energy is said to be stored. 



Elementary Mechanics 33 

Example — If a weight of 200 lbs. were raised 50 feet, 
then £ = 200X50=10,000 ft.-lbs. of work or stored 
energy. 

If the weight were allowed to fall, it would do equally 
as much work, or 200X50=10,000 lbs. In raising the 
weight no account of the time was taken, but when 
allowed to fall gravity determines the velocity in feet per 
second for any period of the fall, air friction not being 

considered. 

V 2 
If in the formula E = Wh, the equivalent of /?, or — , as 

found under Vel. of Falling Bodies, is substituted, the 

formula reads 

V 2 
E—W — , but g = 32.16' and the formula becomes 

* WV 2 



64.32 



Example — If a body weighing 200 lbs. is falling or 
moving at a velocity 50 feet per second, and its motion 
be then arrested in one second of time, the power or 
energy expended would be at the rate: 

,., 200X502 _„,, 

h— . . -. =7773.6 it. -lbs. work per second 
64.32 r 

Now 1 Horse Power = 33,000 ft.-lbs. per minute, or 550 
ft.-lbs. per second, and 7773.6-5-550 = 14.13 HP, the rate 
for that one second of time. 

This formula may also be applied to parts of machines 
moving in any direction, if the weight and velocity of the 
moving parts are known. The principle is observed in 
punching presses, drop hammers, in the shocks incident 
to collisions or in the sudden stopping or starting of 
machines. 



34 Elementary Mechanics 

In the foregoing attention was given to the storing up 
of energy and the power developed when the entire 
motion was arrested during one second of time, and as 
found in trip hammers and the like. In most machines, 
however, only that portion of the stored energy and 
motion is arrested as is sufficient to do the work required 
at any one period. This allows a continuance of move- 
ment in the source of the power and a consequent 
restoring of the energy used, preparatory to it again 
being called on to do work. It will now be seen why 
steam and gas engines, punching and printing presses, 
etc., are provided with flywheels, wherein the energy 
derived from the engine piston or from the driving belt 
is stored and is ready for instant use. Flywheels or 
balance wheels are made relatively large in diameter 
and with a heavy rim, in order to prevent any consider- 
able reduction in speed when power is taken therefrom 
and also to prevent a sudden excessive increase in speed 
after the work is done. The rim represents a succession 
of weights on lever arms (spokes) — equal to a radius 
extending from the center of the shaft to the center of 
the rim. From the theory of levers, the greater the 
radius and the heavier the rim the less should be the 
disturbance for any change due to energy taken out or 
put in. The energy stored in flywheels, where E = total 
energy of flywheel in foot-pounds, W — weight of rim in 
pounds (spokes and hubs usually neglected), V = velocity 
of the rim in feet per second, (using the mean radius of 
rim to center of shaft), g = 32.16, then 

WV 2 WV 2 

E — —z — or — , or the total energy available if 

the wheel is brought to a full stop, but flywheels usually 
are only partly reduced in speed and the energy taken 



Elementary Mechanics 35 

out would be equal to the difference between the total 
energy due to the full speed minus the energy left at the 
reduced speed. If now the velocity at full speed is repre- 
sented by V\, and the velocity at the reduced speed by Vt, 
and the energy taken as E t , then 

^~ 64.32 

The governing of engines in their power output is 
determined by the speed changes in the revolutions of 
the flywheel — that is, if the speed decreases, more power 
is applied to the piston, and, vice versa, less power is applied 
with an increase of speed, below or above a fixed point 
called the rated speed and on which the horse power of 
engine is based. 

The diameter (Radius X2) of a flywheel is limited 
mainly by its speed, for as the speed increases another 
force begins to act and, if excessive, destruction of the 
wheel would result. This force is known as centrifugal 
force. 

CENTRIFUGAL FORCE 

That force acting in a body rotating about a center 
tending to cause it to move in a straight line but pre- 
vented from so doing by the tie which connects it with 
the center of rotation or with parts adjacent to it. If the 
speed be great enough, the forces in the body which hold 
its particles together will be overbalanced and the body 
will fly to pieces, as in the "bursting" of flywheels, and as 
different materials vary in strength, their speed limits, 
too, would vary. The material most used is cast iron. 

The action of centrifugal force may readily be observed 
by taking a weight or ball suspended from a string (see 
Figs. 13 and 14), the end of the string being held in the 



36 



Elementary Mechanics 



hand. If the ball W be started to swing around in a 
horizontal circle, the faster the speed becomes the higher 
will be the plane of rotation until such speed is reached 
when it will swing in a horizontal circle level with the 



w 






! **L 



FIG. 13. 



u 




hand or point of support. Two things will be evident: 
1st, through the speed the attraction of gravity has been 
overcome, and 2nd, the ball is pulling hard against the 
string and the strength thereof prevents the ball from 
flying into space and in a straight line tangent to the 
circle of rotation. This agrees with laws of motion as 
given before. 

Centrifugal force may be expressed by the formula: 

WV 2 

F = — 5- where F = the centrifugal force in pounds, W y 

weight of the revolving body 

V — Velocity of body in feet per second around the axis 

g — Gravity (32.16) as before 

R = Radius in feet. 
The formula may be written in the more convenient form: 

F-. 00034 \JVRn 2 in which n — number of revolutions 
per minute. 

Example — Flywheel rim weighs 500 lbs., revolutions 
per minute 120,fradius 18" (lj ft.). Find centrifugal 
force. 



Elementary Mechanics 37 

F = .000341X500X1.5X120X120 = 3682 lbs. centrif- 
ugal force. 

The centrifugal force increases with an increase of the 
velocity, and such increase in velocity may be due either 
to an increase in the number of revolutions or to an 
increase in the radius, at same rev. per min. 

In the design of flywheels, a rim velocity of 85 feet per 
second is usually taken as a maximum, as with speeds 
much in excess of this the wheel may burst, due to the 
action of centrifugal force, as such force would exceed the 
tensile strength of the cast iron of which flywheels are 
usually made. 

Any additional metal placed in the rim would also be 
stressed beyond its capacity due to centrifugal force. 

In a flywheel rim, the total centrifugal force results 
from weight of entire rim, its velocity and radius. The 
force "D," tending to tear the wheel in two, would 

_ 0.000341 WRn 2 
3.1416 
As relationship between the diameter and circumference 
of a circle =3.1416. Then Z> = 0.00010854/0*^ 



CHAPTER II 



POWER TRANSMISSION 




BELTS AND PULLEYS 

These are extensively used to transmit power from 
one revolving shaft to another, or to a machine for doing 
work. Most machines and tools in shops and factories 
are driven by them (see Fig. 15-a-b-c-d). 

In the simplest form 
(Fig. 15a) the pulley, d, 
on one shaft is called the 
driver and that on the 
other the driven pulley, Fl6 l5a 

d\. The pulley, d, is fixed on the shaft and is rotated 
thereby. Connecting this pulley with the driven pulley is 
a flexible belt, e, of leather, or other similar material. 
This belt when in position hugs the surfaces of the 
pulleys, and its ends being joined together, makes the 
belt endless. 

Now, if power is applied to shaft of pulley, d, and it 
makes one revolution, then, due to the friction between 
the belt and pulley, any point on the belt will be moved 
through a distance equal to the circumference of the 
driving pulley, d, and in the same period of time. If the 
driven pulley, d\, is of equal diameter, it will also make one 
revolution, except for any slip between the belt and the 
pulleys, as the circumference of the one pulley is the 
same as that of the other. 

It will be readily seen, the principles of levers apply 
in each of the two pulleys. The center of the shaft in each 
case is the fulcrum. 



40 



Power Transmission 



If the driving pulley, d, is one-half the diameter of the 
driven pulley, di, the latter would make only one-half a 
revolution to one revolution of the driver, because its 
circumference is twice that of the driver, d. 

If d = diameter of the driver; d\ = diameter of driven; 
both in the same unit of measurement; n = number of 
revolutions of the driver in a given time and fti = number 
of revolutions of the driven in the same time and 3.1416, 
the relation between diameter and circumference; 

Then in any case 3.1416XnX^ = 3.1416XniX^i 

Example — Driver 12" diameter; No. Rev. Driver 100 
r.p.m. and driven 20" diameter 

Diameter: 3. 1416 X 100X12" = 3. 1416 XniX20" 
3.1416 (1200) =3.1416 (20X*i) 
62.832 m = 3769.9200 
3769.9200 



tti = 



62.8320 



= 60 r.p.m. 



Now, as the factor 3.1416 is common to both wheels, 
it may be eliminated and the formula be expressed in 

simpler and better form — 

dn 
niXdi = nXd or ni = -j- 

d x 

or in the same example as above — 

12X100 



«i 



20 




= 60 r.p.m. 

In case "b" (Fig. 15) the 
belt is crossed, instead of 
open as in Fig. 15a, and 
the pulleys turn in opposite 
directions instead of the 
same direction. 



Power Transmission 



41 



In the case (Fig. 15c) a common factory arrangement 
is shown. Here, it is only necessary to combine the 
processes and to express the formula as follows: 

«(r.p.m.) of engine X^X^2X^4 = ^i(r.p.m. of mach.) 

XdiXdsXds 
In which d, d 2 , d\ are driving pulleys, and 
d\, d 3y d$ are driven pulleys 




Cov/iTER SHAPT 



MH 



This may be expressed — The product of the number of 
revolutions of first driver multiplied successively by the 
diameter of each successive driver must equal the 
number of revolutions per minute of the last driven 
pulley multiplied successively by the diameter of each 
driven pulley. 

Example — Engine rev. per min. n = 80, driving pulley 

1st driven, di=l2 } \ 2nd driver ^ 2 =16", 2nd 

driven, d 3 = 12" 
3rd driver, ^ 4 =12", and 3rd driven, ^5=10"; 
to find the number r.p.m. n\ of machine 
spindle? 

Then by substituting in the formula nX^X^X^4 = 
^1X^1X^3X^5; the r.p.m. and the diameters of the 
various pulleys, the condition would be 




42 Power Transmission 

80X36X16X 12 = niXI2 n X 12X10 

, 80X36X16X12 __. ' , 

and f^ = =384 r.p.m. for the machine. 

If the r.p.m. n and fti are given 
and it is desired to provide a pulley- 
train to satisfy the speed ratio, a 
number of combinations with dif- 
ferent diameters may be used, pro- 
vided the product for each of the 
two sides of the equation are equal. 

Note. — In cases where the speed ratio between any pair of shafts 
is large and the one pulley is small and runs at high speed (see Fig. 
15d), the diameter of the said pulley should be taken as the diameter 
d x or d-f-dt (thickness of belt). This is found necessary frequently 
in electric motor drives — where the motor speed is high and the 
driving pulley small. 

The power transmitted by belting is determined by 
the diameter of the pulley, the width of belt and thickness 
of belt and rev. per minute. 

In belt drives, there is always some slip between the 
belts and the pulleys — this makes them impractical 
where a positive drive, as in timed mechanisms is essen- 
tial. In cases of this kind, either a sprocket chain or a 
gear drive is used. In either ease the pulleys are replaced 
by toothed wheels. In the former a chain is substituted 
for the belt, and in the latter no chain is used but the 
teeth are so shaped that those of one wheel mesh with 
those of the other wheel of the pair. 

Should the drive shaft and driven shaft of a gear wheel 
train be too far apart, toothed idler wheels are fre- 
quently used in place of substituting a sprocket chain 
drive. 



Power Transmission 



43 




SPROCKET CHAINS AND WHEELS (PITCH OF) 

Belts of leather, etc., usually consist of one piece, with 
the two ends fastened together forming an endless band. 

Sprocket chains consist of a series of links joined 
together to form an endless chain. 

In Fig. 16 — "<z," a roller sprocket chain is shown, in 
which a roller is held on a cross bar or pin passing through 
the side links and held thereby. 

P — denotes the pitch or center to center distance of 
the rollers. 

The openings in the chain 
are for the reception of the 
teeth of the sprocket wheels 
on which the chain is to be 
used, and the pitch there- 
fore determines the dis- 
tance from center to center of such teeth. 

In b, Fig. 16 — Two links of a roller chain are shown — 
the broken lines representing the links on a straight line 
and the full lines their relationship when passing over a 
wheel with the consequent shortening of the chain toward 
the inside and a corresponding lengthening on the out- 
side of a circle whose circumference passes through the 
link center. This circle, called pitch circle, is the one on 
which the teeth of the wheel must be laid off and its cir- 
cumference must be divisible by the pitch without a 
remainder, so that the number of teeth and number of 
spaces will be equal, and all teeth be alike. The teeth 
prevent slip between the chain and wheels; consequently, 
for any turning movement of the driver there will be a 
like turning movement of the driven wheel and the 
number of teeth (T) in the driving wheel multiplied by 
its number of revolutions (R) will equal the number of 



44 



Power Transmission 



teeth (7\) in driven wheel, multiplied by its revolutions 
(*i), or 

TXR=T 1 XRi^nd R, = ^^ or 7\ = -^£? 

The pitch circle in roller chain sprocket wheels lies half 
way between the top and bottom of the tooth. The 

diameter of the rim of the 
sprocket wheel (see Fig. 16c) 
at the bottom of the tooth 
is called the base diameter 
and is equal to the pitch di- 
- ameter minus the diameter d, 
of the roller. The outside di- 
ameter equals pitch diameter plus the diameter d, of the 
roller. If n = number teeth, P — pitch of chain, 2X = 
angle due to chord (pitch) on the pitch circle, 

, oxr 360° , v 360 180 
then 2X = and X = -zr~ or 



R6.l6c 




P.D.- 



n 



2n 



n 



and the pitch diameter of wheel = 



sin X 



or 



sin 



180° 



n 



GEAR WHEELS— PITCH 



There are two kinds 
of pitch in gear wheels 
— circular pitch (C.P.) 
(•see Fig. 17), the curve 
distance from center of 
one tooth to the center 
of the next tooth meas- 
ured on the pitch circle, 
and diametral pitch 
(D.P.) which indicates the number of teeth on the wheel 
for each inch of the pitch diameter. 




Power Transmission 45 

The pitch in gears, just as with sprocket wheels, deter- 
mines the relative revolutions of the drive and driven 
wheels, and the same formula TxR= 7\Xi?i holds good 
for gears also. 

PITCH CIRCLE 

The circle whose circumference may be divided by the 
circular pitch without a remainder, into as many parts 
as there are to be teeth in the wheel — such parts consist- 
ing of a tooth and a space. 

The pitch circles of a pair of gears working together 
should be tangent to each other and the teeth of the two 
wheels be accurately formed so that the sides of those of 
one wheel may have a rolling contact with the sides of 
those of the other wheel. 

PITCH DIAMETER 

The diameter of the pitch circle may be found readily 
if the number of teeth, n y and either the circular pitch or 
the diametral pitch is given. If n = number of teeth and 
C.P.= circular pitch, then nXC.P. = circumference of 
pitch circle, and circumference 4-3. 1416 = pitch diameter 

nXC.P. 
or pitch diameter = — — — y— ', 3.1416 here refers to the 

relation between the diameter and circumference of any 
circle. 

Again, if n — number of teeth and D.P.= diametral 
pitch, then n divided by D.P. = pitch diameter; 

71 

or, pitch diameter = n p 

As both these formulas give the pitch diameter, they 
are equal to each other and a fixed relationship between 



46 Power Transmission 

the circular pitch (C.P.) and diametral pitch (D.P.) 
may be found, so that if one is given the other may be 
found as follows: 

* * f ,'f ' = y^V or nXC.P.XD.P. = 3.U\6Xn 
v5.14r.Lo JJ.Jr. 

As n is common to both sides of the equation, it may 

be eliminated and C.P. XD.P.= 3.1416 and 

r p 3.1416 3.1416 

If the gear teeth are on a straight bar and the pitch 
line consequently is also a straight line, the pitch circle 
is considered as of an infinite radius and the gear is called 
a rack. The pitch circle in gears lies between the top and 
bottom of the teeth; the part outside the circle is called 
the addendum or face, and the part inside of the pitch 
circle the dedendum or flank (see Fig. 17). This permits 
the teeth of the two wheels to mesh and the pitch circles 
to become tangent to each other, if the teeth and the 
spaces between them are properly formed, by permitting 
the face or ends of the teeth of one wheel to pass into the 
space between the teeth of the other wheel and for the 
proper distance, and vice versa. The dedendum or flank 
is made greater than the addendum or face to permit 
clearance between the ends of the teeth and the bottom 
of the space. 

In some gears, depending upon the class of workman- 
ship, especially in cast gears, which are not machined 
afterward, a side clearance also is allowed — that is, the 
space is made wider than the tooth. 

PROPORTION FOR TEETH 

The basis for proportioning the teeth is the circular 
pitch (C.P.). The addendum is equal to C.P. -^-3.1416 = 



Power Transmission 



47 



.3183 C.P. and likewise the bottom clearance = C.P.-r- 20, 

(C.P.) (C.P.) 
and the dedendum becomes * * + ' ' ' or 

20 3.1416 

3.1416C.P. + 20C.P. = 23.1416C.P. = 

20X3.1416 62.8320 "' 

and the whole depth of tooth = (.3183 + .3683) C.P. = 

.6866 C.P. 

The outside diameter of gear blank = 

Pitch diameter+(2X.3183XC.P.) or 

Pitch diameter+.6366XCP. 

The thickness of tooth for rough gears = .48 X.C.P., 

and the width of the space = .52 X C.P. The sum of which 

equals IX C.P. 

In first-class finished gears, the thickness of the tooth 

C.P. 

and space, each equal .5XC.P. or — ~ ' 

The center to center distance between shafts = 
Sum of the pitch diameters 



TOOTH CURVES 

Two general systems 
are used for finding the 
proper curves for the 
shaping of the teeth — 
theCycloidaland the In- 
volute are such systems. 
The Cycloidal system 
(see Fig. 18) is based on 
finding a circle, called 
generating circle, which c /cloidal 

is alternately rolled on the [inside and outside of the 




48 



Power Transmission 



pitch circles with a trace point, which describes two 
curves for each wheel, an Epicycloid Curve for the face, 
and a Hypocycloid Curve for the flank. 

The diameter of the generating circle is found by taking 
one-half the diameter of a 12-tooth pinion and of the 
pitch being used (see Fig. 18). A generating circle based 
on the 12-tooth pinion generally is used, for with it the 
flanks of the teeth become radial. 



THE INVOLUTE SYSTEM 

In this system (see Fig. 19) a line drawn at an angle of 
75§° to the line of gear centers defines the angle of 
approach of the teeth of the two wheels. This line 

crosses the line of centers 
at the point of tangency 
^ .. of the two pitch circles. 
A circle inside and concen- 
tric with the pitch circle 
drawn tangent to the line 
of contact is called the 
base circle. An involute 
curve starting from the 
point of intersection be- 
tween the base circle and 

INVOLUTE. i« e r i 

— — line of centers lorms the 

curve for the addendum, or face, and part of the deden- 
dum, or flank, of the gear tooth, respectively, of each 
wheel. The balance of the flank, or to the root of the 
tooth, is drawn radially toward the center of the gear, 
starting at the base circle. 

Cycloidal teeth are considered weaker than involute, 
and for good operation the pitch circles must always be 




Power Transmission 49 

tangent one to the other. Also 24 cutters of each pitch 
are required to cut gears ranging from 12 teeth to a rack. 

Involute teeth are considered stronger and the pitch 
circles need not be exactly tangent one to the other for 
good operation. Again, only 8 cutters are required to cut 
gears ranging from 12 teeth to a rack, nor is side clearance 
usually allowed in cut gears, the tooth thickness being 
equal to one-half the pitch. 

Involute teeth are used more extensively for the above 
reasons and practically universally so for bevel gearing. 
In the involute rack the sides of the teeth are straight 
and at right angles to the line of contact, except that the 
tops are often rounded off to prevent interference. 

SPUR GEARS 

In gear drive k where two shafts are parallel and the 
teeth throughout their length are parallel to said shafts, 
the gears are known as Spur Gears. 

BEVEL AND MITRE GEARS 

In drives where the shafts are not parallel and where 
their center lines, if projected, would intersect, are 
classed either as bevel or mitre gears, depending upon the 
angle of intersection and number of teeth in the mating 
gears. If the angle is 90° and both wheels have a like 
number of teeth, they are called mitre gears. 

In spur gears the pitch circles, etc., are considered at 
the end of a short cylinder. In bevel and mitre gears 
they form the bases of the frustums of pitch cones whose 
apexes meet at the point of intersection of the two center 
lines (see Fig. 20). 

In the illustration the pitch diameters L and S, re- 
spectively, form the bases of the pitch cones. The line of 



50 



Power Transmission 



FIG. 20. 




FIG. 20 B. 



FIG. 20C 



Power Transmission 51 

tangency of the two cones is the pitch line, because at any 
point throughout its length it would define the relative 
diameters of the gear and pinion at that point. 

The pitch diameters and number of teeth, as in spur 
gears, determine the speed ratio between the two shafts. 
The back or cutting face (see Fig. 20) of the gear and 
pinion, respectively, is at right angles to the pitch line 
and at the point of tangency of the two pitch diameter 
circles. On this face, above and below the pitch line, the 
addendum and dedendum of the teeth of each gear are 
laid off and in the same proportions as for spur gears. 
In like manner, the thickness of the teeth at cutting face 
would be the same as for spur gears. The teeth in each 
wheel would, however, decrease in size toward the point 
of intersection of the two shaft centers. 

As before stated, the involute system is used almost 
universally for bevel gears. To develop the tooth curve, 
the pitch diameters L and S would serve well enough for 
the thickness of the teeth and width of space between 
them, but as they are at right angles to their respective 
shafts the addendum and dedendum for each wheel 
would be foreshortened, and the true tooth curve not be 
obtained. If, however, the development of the tooth 
curve be on a plane at right angles to the pitch line and at 
the point of tangency of the two pitch circles, said plane 
(cutting face) is common to both gears and the addendum, 
dedendum and thickness would show in their relative 
proportions. The cutting face of each, however, (see 
Fig. 20b) forms the side of a cone each having as its base 
the pitch diameter and a slant height R or r\ depending 
upon which gear is considered. To develop the surface 
of a cone (see Fig. 20c); the slant height ri is used as a 
radius and an arc drawn with a length at the circumference 



52 Power Transmission 

equal to that of the circumference of the base or S X 
3.1416. 

The developement (Fig. 20c) is of the same dimen- 
sions as the back cone of the small gear. 

If then the part circumference of the development is 
taken as equal to the length of the -pitch circle, and 
divided into as many spaces as the gear has teeth, and 
the teeth be laid off as in spur gears, then the cone formed 
by the development would coincide with the small gear, 
with the teeth of proper form and number and the base 
at right angles to the center line and of the right pitch 
diameter. 

As it is unnecessary to develop more than one or two 
teeth on each wheel, the method shown in Fig. 20c is 
generally combined with the general drawing, or as shown 
in Fig. 20, wherein the line of centers is formed by R 
and r\ and the line of contact would pass through the 
point of tangency of R and r\ at an angle of 75|° to the 
line of centers formed by them. 

The lines forming the tops of the teeth and usually 
those forming the bottom of the teeth, all converge to 
the apex or point of intersection of the cones. 

WORM GEARING 

Worm gearing is used to transmit and convert the 
power of a fast rotating shaft driven by a small force 
moving at a high velocity to a slow rotating shaft to 
produce a great force moving at a low velocity. It is 
smooth in its action and quiet. Good lubrication is 
essential to its life and also to reduce the loss of power 
due to transmission. 

Worm gearing consists of a screw, the threads being 
on the exterior of a short cylinder (see Fig. 21) mounted 



Power Transmission 



53 



FIG.gl. 



on the drive shaft. The threads are similar to the involute 
teeth of a rack. The screw thread engages screw threads 
on the edge of a disc, called the worm wheel fixed on the 
driven shaft in order to rotate it (see Fig. 21a). 

The worm wheel may be likened to the 
half side of a continuous nut, the threads 
of which are repeated aroundthe entire cir- 
cumference of the worm wheel disc. By this 
means a continuous turning of the worm will cause also 
a continuous turning of the worm wheel but at a reduced 
ate, depending upon lead of the worm. A worm may be 
provided with one or more threads parallel 
to each other and be known as a single, 
double, etc., thread worm. The longitudi- 
nal distance from any point on a thread 
to a like point on the same thread after 
passing once around the worm cylinder is 
called the lead, and determines the move- — 
ment of the worm wheel for one revolu- 
tion of the worm. The longitudinal dis- 
tance from a point on a thread to a like point on the next 
thread is the pitch. 

Therefore with a single thread worm (Fig. 21c) the 
pitch equals the "lead" and one revolution of the worm 
advances the worm wheel 
one tooth. With a double 
thread worm (Fig. 21b) the 
pitch equals one-half the 
lead and for each revolution the worm wheel is advanced 
two teeth. With a treble thread worm the pitch equals 
one-third the lead and for each revolution the worm 
wheel is advanced three teeth. The pitch represents the 
distance from center to center of the teeth of the worm 
and worm wheel. 




FIG.2IA 




SINGLC 



Fi6.2ia 




FIG.2IC 



54 Power Transmission 

If T= number teeth in the worm wheel; ni = number 
revolutions of the worm wheel, WL = lead of worm, and 
WP = the pitch of the worm, both in the same unit, and 
n — number revolutions of the worm. 

_, Lead (^Z)XRev. (n) 

1 nen n\ — 



Teeth (T) X Pitch (WP) 

Example. A single thread worm makes 120 r.p.m. and 
has a pitch and lead of 1 inch. The worm 
wheel has 48 teeth. How many revolutions 
per minute will the worm wheel make? 

120X1 o1 

Example. With the same example as above, but with a 
double thread worm; worm 120 r.p.m., pitch 
1", lead 2", worm wheel has 48 teeth; to find 
r.p.m. of worm wheel? 

120X2 B 

* i= wr =5r - p - m - 

Example — Same as above except worm has treble thread: 
120X3 _ 
Wl = "48xT = 7 ^ np - m - 

The speed ratios, in the three cases are: 
120 to 2\ or 48 r.p.m. of worm to 1 of worm wheel 
120 to 5 or 24 r.p.m. of worm to 1 of worm wheel 
120 to 7 J or 16 r.p.m. of worm to 1 of worm wheel 
Now, if in each case the force acting to turn the worm 
were the same, the force which can be exerted by the 
worm wheel, under the respective speeds 2\ — 5 — 7j 
r.p.m., would be respectively 48 — 24 — 16 times as great, 
neglecting friction, which plays, however, an important 
part in gears of this type. 



Power Transmission 



55 



Worm gearing is extensively used in drives for eleva- 
tors, automobile trucks, mechanical stokers for boiler 
furnaces, the guiding mechanism of traction engines. 
The two shafts are usually at right angles to each other 
and always on different planes. The worm or worm 
shaft must be furnished with an end thrust bearing, as 
the action of the worm is similar to that of a screw jack. 

CAM 

A cam is a device used extensively in machines, such as 
steam and gas engines, printing presses, automatic ma- 
chinery, sewing machines, etc., for changing the direction 
of a motion or to produce either an intermittent or a 
graded motion. A cam transmits its motion, or a part of 
it, to a follower which is moved thereby and in the changed 
direction. A considerable loss from friction occurs in 
cam drives. The follower preferably should be provided 
with a roller, to engage the cam surface. The action of 
a cam is to engage and push the follower out of its way. 

This path of movement may be straight, as in a re- 
ciprocating cam, or curved as in rotating edge cams and 
rotating cylinder cams. (See Fig. 22, a-b-c, respectively.) 

The follower may be arranged on a slide as at <z, or on 
a lever arm as in b, and c. 



r ±g 





Types — Bar, disc and cylinder cams. Fig. 22a, rep- 
resents a reciprocating bar provided with a cam projec- 
tion at one side, having an inclined face to engage 
the shoe of the follower. At b (Fig. 22) an edge cam 



56 Power Transmission 

is shown with the cam projection or incline built on the 
outside edge of the disc. 

In Fig. 22c a cylindrical cam is shown in which, 
instead of the cam projection, its counterpart, a cam 
groove is cut into the face of the cylinder to engage the 
follower. 

In any case the principle of the inclined plane comes 
into play. The force exerted being the product of weight 
X height in feet lifted equals the foot-pounds of work — 
friction not considered. The approach to lifting face of 
the cam is usually curved to reduce the shock, incident 
to the meeting of the cam and follower. 

EFFICIENCY 

The percentage of useful work obtained from any 
machine or process for the known power exerted. The 
basis is 1 or 100 per cent — so that if 100 ft.-lbs. of power 

90 

is required to do 90 ft.-lbs. of useful work, then t^t: = -90 

or 90 per cent, is the efficiency of the machine or process. 
Friction is one great obstacle to efficiency in any machine 
or process; lubrication therefore, in machines saves power 
and helps toward efficiency. 

All machines and processes are more or less efficient, 
and the greater the useful return for power expended the 
higher would be the percentage of efficiency. 



CHAPTER III 




TRIANGLES 

Their Use and Solution — 
Any figure bounded by three 
straight sides as A-B, B-C, 
C-A, joined at their ends 
forms a triangle. Inside the FIG.23 /\ 

figure (see Fig. 23a) an angle is formed between each side 
and the side adjacent to it, or three (tri) angles a, b, c, 
as the name triangle implies. A triangle consists of six 
parts as follows: 3 sides A-B, B-C, C-A and 3 angles a, 
b, c. If three of the parts are known the other three may 
be found, provided one known part is a side, otherwise 
we could find only the proportionate relation of the sides 
but not their numerical value. 




An angle is defined by the divergence, expressed in 
degrees, between the two sides forming it. If any line is 
taken as a radius and caused to make one revolution, it 
would describe a circle or would have passed through 
360 spaces called degrees (°). This is true whether the 
line be short or long(see Fig. 23B). It may also be readily 
observed by comparing the dial of a watch with that of a 
tower clock — the minute hand in each covers the 12-hour 



58 Triangles 

space in one revolution. A circle, therefore, contains 
360°; and for purposes of making more exact computa- 
tions each degree is divided into 60 parts called minutes 
(') and these again subdivided into 60 parts called 
seconds ("), so that a circle contains 360° (degrees) or 
21,600' (minutes) or 1,296,000" (seconds). 

In one-half circle (semi-circle) A to C, there are 180°, 
and in a quarter circle or right angle (from A to B) there 
are 90° (see Fig. 23C). The sum of the three angles in a 
triangle is equal to 180°. Proof: In the triangle A, B, C, 
side A-C is also the diameter of the circle with D as the 
center. AD, BD and DC are equal, but BD is at right 
angles to both, so that AB and BC must be the same 
length. If the left triangle ABD is folded over along the 
line BD, AD would cover the line DC, and AB would 
cover the line BC, showing that the angle: a — angle c and 
angle b = angle b\ As b and b ] are equal to each other, 
each must equal one-half the angle formed by ABC. 
Again, if the triangle ABD is revolved about D, letting 
AD rest on BD, then BD would rest on DC and AB on 
BC, showing that angle a also equals b } and b equals c; 
but as previously found, a also equals c; therefore a also 
equals b. Again, the same triangle ABD, if revolved 
about B so that BD takes the position BDi parallel with 
DC, and AD the position AC parallel with BD, AB will 
fall on the line BC, a will be on one side and c on the other 
side of the line BC and the sum of the two would equal 
the angle DCDi. If, however, the triangle ADB is placed 
over BDC with AD on DC, then BD would fall on D X C, 
and as ADB = 90° so will c + # = 90 o , but a and c being 
equal, each will be 45°, so also £ = 45° and b-\-bi = 90° or 
the total number of degrees in triangle =45+90+45 
or 180°. 



Triangles 59 

The sum of the angles in all triangles is equal to 180°, 
so that if two angles of a triangle are known, their sum 
deducted from 180° will give the number of degrees in 
the third or remaining angle. 
Triangles are of three kinds: 

Acute angled, when the angles are less than 90° 

Right angled, when one angle is 90° 

Obtuse angled, when one angle is more than 90° 

(See Fig. 24 — a, b and c.) 
In computations involving the solutions of triangles, 
certain relationships (called trigonometrical functions) 
have been established: they are based on the relationship 
existing between the sides and angles of right angle 
triangles. 






FIS.24C 

Problems embracing right angle triangles where the 
lengths of two of the sides are known may be solved 
readily, as — ___ 

The (hypothenuse) 2 = the sum of the (base) 2 plus the 
(altitude) 2 , or as usually expressed: 

The square of the hypothenuse equals the sum of the 
squares of the other two sides. (See Fig. 24 — b.) 

The relationships of the three sides are given by the 
formulas: 

Wherein m is the hypothenuse, o the base, and n the 
altitude. 

m 2 = n 2 -\-o 2 

m = Vft 2 + o 2 , n— ■\m 2 — o 2 and o= ■\m 2 — n 2 , 

Example. m = 5, n = 3, angle c = 90°. To find o. 

Solution. o 2 = m 2 -n 2 oro 2 =(5X5)--(3X3)=25-9 = 16. 
o= Vl6 = 4. 



60 



Triangles 



FUNCTIONS OF RIGHT ANGLE TRIANGLES 

(See Fig. 25.) They are known as follows: 

Sine (abbreviation sin) — The side opposite the angle 
when the hypothenuse is considered as the radius of the 
circle. 

Cosine (abbreviation cos) — The side adjacent to the 
angle when the hypothenuse is considered as the radius 
of the circle. It is the sine of the complementary angle, 
i. e., the sine of the angle remaining, after deducting the 
angle, first mentioned from 90°. 

Tangent (abbreviation tan) — The side opposite the 
angle when the base is considered as the radius of the circle. 

Cotangent (abbreviation cot) — The side opposite to the 
complementary angle with the base as the radius of the 
circle. 

CQTAN . - 




^ —COSINE — H 

FIG. 25. 

Secant (abbreviation sec) — The hypothenuse of the tri- 
angle when the base is considered as the radius of the 
circle. 

Cosecant (abbreviation Cosec) — The hypothenuse of 
the complementary angle when the base is considered 
as the radius of the circle. 



Triangles 



61 



The sines and cosines of angles lie wholly within the 
arc of the circle. 

The tangents and cotangents of angles lie wholly with- 
out the circle and are tangent thereto as the names imply. 

The hypothenuse of the triangle, when dealing with 
sines and cosines, acts as the radius of the circle, but when 
dealing with tangents and cotangents it is the line which 
connects either the end of the tangent or the end of the 
cotangent with the center of the circle, with the base of 
the triangle considered as the radius. In either case 
part of the secant or cosecant lies within and the balance 
without the circle. 

The functions of angles indicate the relationship or 
ratio between the sides of right angle triangles and by 
their use the actual lengths of the sides may be found, 
provided three parts of the triangle are known, one of 
which is a side. 

Tables of the functions of right angle triangles, called 
: 'Tables of Trigonometric Function," may be found in 
the various handbooks. In order that the student may 
understand how to apply such 
tables and how the approximate 
values may be found, attention 
is called to Fig. 25A, etc, in which 
let, R indicate a radius of any 
length, and its value assumed 
as the whole number one (1) or 
100%. Now by drawing an arc 
of 90° and forming a right angle 
between the two extreme posi- 
tions the percentage value of sines and cosines may be 
readily obtained, if the arc be divided into 90 equal parts 
each representing a degree (°) and the two sides, base and 





62 Triangles 

altitude, each be divided into 100 equal parts, each part 
being 1-100 or 1% of the whole or radius (1). In the 
diagram for the sake of clearness only the 15° intervals 
and the 10% distances are given. For accurate computa- 
tions, the trigonometric tables only should be used. 

Assuming as an example (see Fig.,25B) a right angle 

triangle with the hypothenuse equal to 15" and the angle 

confined between it and the base, a, be 

30° and the angle between the base, a, 

and the altitude, b, of course 90°. To 

FIG.25B . find the respective lengths of the side 

opposite, or b, and the side adjacent, or a, to the 30° 

angle. 

Now, by referring to the diagram, R would represent 
the hypothenuse, or 15", moved to the position indicated 
by the broken line and meeting the arc at the 30° divi- 
sion, the side opposite and the side adjacent to the 30° 
angle would bear the same percentage relationship to the 
hypothenuse and to each other as the base and altitude 
of the given triangle bear to the 15" hypothenuse and to 
each other. If now the altitude x, be measured or read, 
its length will be found to be equal to 50 parts, or 50%, 
of the hypothenuse, and in the case of the 15" hypothe- 
nuse with the 30° angle the altitude would be 50%, or 
one-half of 15", or 7J". If, in like manner, the distance y, 
side adjacent be measured, it would be found to be nearly 
87% (actually .86603) of the radius or hypothenuse, and 
if the 15" be multiplied by .86603 the product or 12.9905" 
would be the length of the base or side adjacent to the 
angle of 30°. The same method would have to be used 
for any angle when considering sines and cosines; the 
sine being the side opposite or altitude, and the cosine 
the side adjacent or base of the triangle. In the above 



Triangles 63 



o 

5 



example two of the angles were known, i. e., 30° and 90 
total 120°, have been accounted for; by subtracting this 
sum from 180° the value of the third angle or 60° also is 
found. The value of all three sides and all three angles 
now are known. The 60° angle in this case is the comple- 
mentary angle, as it represents the difference between the 
90° and 30°. If the triangle were turned so that the side, 
y, represented the altitude, it would still be of the same 
length, but would be opposite the angle of 60° and there- 
fore be the sine of 60°; in like manner, the side, x, would 
be the cosine of 60°. It will now be understood that a side 
of a triangle may be considered either as a sine or as a 
cosine, depending which of the two sides of the right 
angle is taken as the base. It will also be seen that the 
sine of an angle is equal in value to the cosine of the com- 
plementary angle, and vice versa; therefore the affixing 
the co to sine to indicate the sine of the complementary 
angle. 

By examining a table of natural sines and cosines, it 
will be observed that the sines increase from to 100% 
or 1, while the cosines decrease from 1 or 100% to 0, 
in passing around the arc from 0° to 90°. 

The tables usually are so arranged that when finding 
a sine you follow the degree readings in the column 
(1st col.) and the sine column (2nd col.) from the top 
downward to 45°; then jump to the 3rd column for sines, 
and the last column for the degrees, reading upward 
until 90° is reached. Tables are more complete and more 
accurate than the diagram, and therefore are used in 
preference to the diagram. 

In cases where the sides of the right angle triangle are 
given and it is desired to find the angles confined between 
the sides, it is necessary only to divide the side opposite 



64 Triangles 

the angle to be obtained by the hypothenuse and find 
the angle in the table under sines corresponding to the 
decimal obtained from such division. If, on the other 
hand, the side adjacent had been taken as the dividend, 
then the decimal equivalent would have to be sought for 
under the head of cosines in the table and the angle in 
degrees, etc., taken found opposite thereto. 

TANGENTS AND COTANGENTS 

These are used relatively in the same manner as the 
sines and cosines, but differ in magnitude due to being 
wholly without and tangent to the arc. 

SECANTS AND COSECANTS 

When solving right angle triangle on the basis of 
tangents and cotangents the secant and cosecant repre- 
sent the hypothenuse, as already stated. The prefix co 
is used for the same purpose as with sines and tangents. 



CHAPTER IV 



FLUIDS 

Those substances such as gases and liquids whose 
particles have little or no attraction for each other and 
may be easily separated. All fluids have weight; they 
tend to seek their level and are affected by heat and cold. 
The gases, for example air and steam, may be compressed 
by pressure, and likewise will expand with a reduction in 
pressure. The liquids, for example oil and water, are 
relatively incompressible. Air and water are the most 
widely known fluids. 

Air, subject to excessive pressure and cooled, may be 
changed from a gas to a liquid (liquid air). 

Water readily assumes three states, due to changes in 
temperatures: 

Gaseous (steam) above 212° F. 

Liquid (water) between 212° and 32° F. 

Solid (ice) below 32° F. 

It may be possible that air, too, would assume a solid 
state if the pressure were sufficient and the temperature 
low enough. 

The whole earth's surface is covered by a blanket of 
air, which, due to its weight, exerts a pressure of 14.7 
pounds on each square inch of surface at sea level. On 
the tops of high mountains the pressure is correspondingly 
less, due to a less thickness of the blanket. 

Water (fresh) weighs approximately 62.5 lbs. per cu. ft. 

Water (salt) weighs approximately 64 lbs. per cu. ft. 

As a cubic foot contains 1728 cu. in., then 62.5 -5- 1728 = 
weight of 1 cu. in. or a 1 M cube, and the pressure (weight) 
exerted by a vertical column 1" sq. in.X12 inches high 

would be — ' =62.5-7- 144 = .434 lbs., nearly, pres- 

sure per sq. in. for every ft. in height. 



66 



Pressure 



As air exerts a pressure of 14.7 lbs. per sq. in. at sea level, 
then 14.7-^.434 = 34 ft., the height of a column of water 
which balances approximately the atmospheric pressure. 

Fluids have different weights and deusities, which 
explains why oil floats on water, also why a balloon filled 
with a gas lighter than air rises — also why heated air 
rises and the cause of a draft in a chimney. 

Fluids when confined under pressure, whether due to 
weight pressing down or otherwise, press against all sides 
uniformly at right angles to such surface and for each 
square inch thereof. Solids press only in the direction 
of the applied force producing the pressure. A block of 
ice below freezing temperature would, like wood or stone, 
transmit an applied force in a direct line to the surface 
upon which it rests without outside aid, due to the strong 
attraction of its particles (molecules) for each other. 

If the temperature be raised above freezing, the ice 
would melt and run off in the form of water and allow 
the weight to settle, because in melting the strong molec- 
ular attraction is destroyed and the molecules try to 
get from under the load and seek their level, whether 
the load be an extraneous weight or superimposed ice or 
water. If, however, the block of ice had been fitted neatly 
i • into a vessel or cylinder and then 

! It ' 

melted, the walls of the vessel would 
prevent the water from running off 
and hold the molecules to their work 
against their normal desire to get 
away in any direction, if free to 
move upon each other. If the cyl- 
inder, A, as in Fig. 26, had been 
provided with a neatly fitted slid- 
ing piston resting on the water or melted ice and sup- 
ported a weight, w 9 the force due to such weight would 



», s 

r— s*. 



54 



E 



& 



w 



in 



Fig.ge 



Pressure 67 

be distributed evenly over the entire water surface and 
the downward pressure on each square inch would be 
resisted by an equal upward pressure per square inch 
under the law of balanced forces. The molecules forming 
the water, being restrained by the cylinder against move- 
ment, transmit the force from one to another in all 
directions alike and finally to the cylinder walls and at 
right angles thereto, which react with the same intensity 
per unit of surface as the force due to the weight applied. 

Had the piston been made very loose fitting, so that 
the water could readily pass between it and the cylinder, 
a condition similar to that of the non-restrained melting 
ice would occur, because the molecules could get from 
under the load and be pushed up alongside the piston 
and allow it to settle or sink. The opening of a valve or 
stop-cock, as at a> would produce a like result. 

If, in Fig. 26, a second cylinder, B, with a neat fitting 
piston be connected with the first by a pipe and valve, c, 
and said valve be opened, the pressure per unit area in 
the two cylinders would be equal, but the total pressure 
or weight supported would be to each other as their 
respective areas, or as to the squares of their diameters. 

Example. A weight of 50 lbs. is on the small piston of 
2" diameter; what weight will a large piston 
6 M diameter sustain if the cylinders are 
connected? 

Let. d — diameter small piston 

and d\ — diameter large piston 
W — weight on small piston 

W\ — weight which can be supported by large pis- 
ton 

tu w WXd? 50X36 1800 .__„ 

Then Wi= « — = — j— = — r— =450 lbs. 

a 2 4 4 

If weight, W\, is to be lifted 1 inch in the above exam- 
ple, the weight, W, on small piston would have to descend 



68 Pressure 

9 inches (i. e., ratio of pistons) — as the inch-pounds of 
work are the same in either case. 
/FX9" = /F 1 X1" 
50 lbs. X9 M =450 lbs.Xl" 

The above is the principle of the hydrostatic press for 
exerting very heavy pressures over limited distances. In 
the place of one long stroke of the small piston, the small 
cylinder is arranged as a force pump. 

If both pistons were removed from the cylinders, the 
water columns would balance each other and their sur- 
faces be level. Hence the expression "water seeks its 
level." The vertical distance from the center of the con- 
necting pipe to the surface of the water or the head, 
h, would be the same, and if said pipe were 1 square 
inch in area (approximately lf'dia.) and the head, ^ = 5 
feet, then the opposing pressures, P, in said pipe in 
lbs. per square inch from each cylinder would be, 

P = h in feet X. 434 (weight of column of water 1 foot 

high and 1 square inch in cross section) 
or P = 5X.434 = 2.17 lbs. per sq. inch 

In Fig. 26, if the small cylinder A were extended, as 
indicated by the broken lines, and the weight and piston 
removed and replaced by 50 lbs. of water the large weight 
(450 lbs.) would be balanced, as before; the water, surface 
in the small cylinder would, however, be proportionately 
higher than in large cylinder. This height may be found 
as follows: 

If pipe = 2"; its area equals 3.1416 square inches 
and W— weight of water equals 50 lbs. 

Then 50-f-3.1416 = lbs. water for each sq. in. of surface 
but .434 lbs. equals weight of 1 sq. in. column of water 
1 ft. high, so that the height (hi) is 

W+diam. 2 X. 7854 -t- .434 = h x 
or 50-^-3.1416-f- .434 = 36.7 ft. 



Pressure 



69 



In tanks the shape determines only the volume and the 
consequent total weight to be supported. The height of 
water or head from the bottom of the tank to the water 
surface determines the pressure on each unit of surface 
of the bottom, and is independent of the volume. The 
unit pressure on any intermediate plane will be propor- 
tional to its depth below the water surface. 

Figs. 27-A-B-C-D-E show tanks of various shapes, in 
which d is the large diameter and d x the small diameter; 
h is the total height in feet from the bottom to the water 
surface, and hi the height in feet from the water surface 

to b, h. 

The volume of A is the largest, the vol. B = vol. C, and 
vol. D = vol. E, but each is less than A, and likewise the 
weight of the water contained is less. 

-Jftw- 




The diameters, d, are equal in each case; so, too, the 
diameters, d\, are equal to eachjDther. 



70 Pressure 

The head, A, in feet is the same in each case 
Also the head, hi, in feet is the same in each case. 

If P represents the unit pressure in lbs. per square inch 
at the bottom and Pi the unit pressure in lbs. per square 
inch at the point b, b\, 

Then P = AX.434 lbs. 
and Pi = AiX.434 lbs. 
Exam-pie. If A = 5 ft. and hi = 2.5 ft. 

d=12 u diameter (113.1 sq. in. area) 
^i = lf" = dia. (1 sq. in. area). 

Find the weight of water contained; the 
unit pressure per square inch and total pres- 
sure on the extreme bottom in A, B, C, D 
and E, respectively. 

Also to find unit pressure and direction on 
b, b and bibi of B and C, respectively. 

A contains 6786 cu. in. -M 728 cu. in. per cu. ft. X62.5 
lbs. per cu. ft. = 245.4 lbs. 

B and C each contain for 

large part § of 245.4 lbs. or 122.7 lbs. \ 124.7 

small part 54 cu. in. -M 728X62.5 or 1.95 lbs. J lbs. 

D and E each contain 2466.2 cu. in -M 728X62.5 = 

89.2 lbs. 
The pressure per square inch on the bottom of A, B, 
C, D, E, respectively, is 5 ft. X. 434 lbs. per sq. in. for 
each ft. =2.17 lbs. 
and the total pressure on the bottom of 

A =113.1 n,, X2.17 = 245.4 lbs. pressure 

B =113.1 X2.17 = 245.4 " 

C = 1. X2.17= 2.17 " 

D =113.1 X2.17 = 245.4 " " 

E = 1. X2.17= 2.17 " 



Buoyancy 71 

The unit pressure per square inch on b, b, and hi, b\ is 
2.5' X. 434 or 1.085 lbs., and the total pressure outside 
the small column 

113.1 aM -l° M =112.ia n X1.085 = 121.53 lbs. 

In the case of B the pressure is upward, and in C the 
pressure or weight is downward. 

In B and D the total pressure on the bottom is the same 
as in A, due to the reactions of b and b^ and full head, h y 
as in A. The walls hold the top and bottom plates from 
separating. 

In C and E the total pressure at extreme bottom is the 
same, due to the upward reactions of b\ and £ 3 , respec- 
tively. 

BUOYANCY 

The relative lightness of two substances, one or both 
of which are fluids. The lighter substance is buoyed up 
and floats, and any substance tends to float in the fluid. 

A block of wood 1 cu. ft. or 45 lbs. will 
float in water, a part being above and the j] 
balance below the water surface. A block 
of iron 1 cu. ft. or 450 lbs., however, will 
sink to the bottom. (See Fig. 28.) 

The weight of water displaced by that 
portion of the wood block below the sur- 
face will equal the total weight of the block of 45 lbs., 
which is only 72% (45-^62.5) of the weight of an equal 
volume of water. Each, however, has the same area 
and height, so that for the same weight the wood block 
would be immersed only 72% of its height or .72 ft. (8.64 
inches). 

The block pushes down with a force of 45 lbs. on 144 
sq. in. of surface, and must be pushed up with the same 
force, or 45 -f- 144 = .3 125 lbs. per sq. in. Water per ft. 



^r 






2 



Fie.ge. 



72 BOUYANCY 

height exerts .434 lbs. pressure per sq. in. (see water, 
under fluids). Then .3 125 -r- .434 = .72 ft. (8.64 inches), 
the depth the block is immersed to provide the requisite 
pressure of .3125 per sq. in. to support it. 

Now, the iron block being totally immersed with 
water above, it is pressed down by a force equal to h 
(head in ft.) X. 434 lbs. per sq. in. X area (144 sq. in.), 
but is also pushed up from underneath by a force AiX 
.434X144 sq. in. 

It will be seen that, due to the greater head, the force 
from underneath will be the larger and tend to lift the 
block- — the difference in their unit weights (62.5 water 
and 450 lbs. iron), however, is too great. 

Example. A block of cast iron 12 M X12 M X12" weigh- 
ing 450 lbs. is placed in a vessel of water. The depth from 
water surface to bottom is 4'-0". How much greater is 
the force pushing up than that pushing down? 

Solution. 4'-0"X.434 lbs. per sq. in = 1.736 lbs. pres- 
sure per sq. in upward; 3 , -0 M X.4'34 lbs. per sq. in = 
1.302 lbs. pressure per sq. in. downward, or .434 lbs. 
less; then 144 sq. in. X. 434 = 62.496 lbs. excess of 
pressure upward — but this equals the weight of the 
1 cu. ft. of water displaced by the 1 cu. ft. of iron. 

Now, if the block of iron 
were weighed in the water, as 

| 367. 5 LPS . 

shown in Fig. 28A, it would 
weight 450 - 62.5 lbs., or 387.5 
lbs. 

It may be shown that the shape of the block or the 
depth to which it is sunk does not alter the above if the 
volume remain the same, because for each cubic foot 









*' JL 


i=~ 




=~— 


G^ 








== 


IRON 


H 


FIG.30A. 


— - 


— 


'=£_-= 




: i-rr 





Buoyancy 73 

of iron or part thereof a like volume of water would be 
displaced and the buoyancy per unit of weight remain 
the same. 

If the volume of the immersed substance be increased, 
the weight remaining the same, the upward tendency 
would increase in proportion to the increased displace- 
ment. In like manner, if the volume remain the same 
but the weight be decreased, the upward tendency would 
increase. 

Example. Will a hollow cast-iron cube 12" X 12" X 12" 
with | inch thickness of wall float in water if cast iron 
weighs .2607 lbs. per cu. in.? 

Solution. 1 cu. ft. of water weighs 62.5 lbs. 

Cube — 4 sides XI If" wide = 

47X11| M high = 540.5 sq. in. 

Top and bottom 12 M X12" = 

144X2 = 288. sq. in. 

825.5XJ M =207.1 cu. in. 

207.1 X. 2607 lbs. per cu. in. = 53.991 lbs. weight of 
cast-iron hollow cube. 

62.5 lbs. weight of cube of water — 54 lbs. weight of 
cube of iron = 8.5 lbs. iron cube is lighter than 
displaced water. 

The cast-iron cube will float, therefore, and would 
sustain approximately 8| lbs. more than its own 
weight. 

It will now be seen that steel vessels — even though 
steel is heavier than water — can be constructed to be 
capable of displacing an amount of water exceeding 
their own weight, and thus be suitable for cargo purposes. 



74 



Pumps 



PUMPS 

Machines for raising water or other fluids from either 
a low elevation or a low pressure to a higher elevation or 
pressure, and usually understood to refer to those 
machines wherein one or more movable members (pistons, 
plungers or impellers) act directly on the fluid to be 
moved. 

Fluids (gases and liquids) always flow from a state of 
high pressure to a lower one. 

Pumps are known as of two general types: 

Lift Pumps — Those for lifting water usually under 
atmospheric conditions from wells or cisterns to the 
surface of the ground or operating platform. (See 
Fig. 29A.) 
Force Pumps — Those whereby the water is dis- 
charged against considerable pressure, whether due 
to differences in elevation or differences in pressure, 
as in feeding steam boilers. (See Fig. 29B.) 

In Fig. 29A a simple lift 
pump is shown, and in Fig. 
29B a simple force pump is 
shown. Each consists of a 
pump cylinder, a; a recip- 
rocating piston or plunger, b; 
a suction pipe, c, leading to a 
point below the water surface, 
d; a suction valve, e, to per- 
mit water to flow from the 
suction pipe to the cylinder, 
but not to return; a plunger rod,/, fixed to the plunger 
at one end and pivoted to an operating lever, g, at the 
other end; the discharge outlets, A, hi, respectively — the 
one h, at the top of the lift pump cylinder and above the 




FI6.29A 



F1Q.23B- 



Pumps 75 

plunger; the other, hi, at the bottom of the force pump 
cylinder and below the plunger; discharge valves, k, k\, 
respectively, are provided — the one, k, for the lift pump, 
and the other, ky, for the force pump; a passage, /, through 
the lift pump plunger is controlled by the valve, k, which 
permits the water to flow upward through the plunger 
but not return. 

It will be noticed that each type has the same number 
of parts, the differences being only in the discharge out- 
lets and valves, and that one plunger has a passage 
through it while the other is solid. 

The plungers of pumps are arranged to slide freely in 
the cylinders, but sufficiently tight to prevent water from 
passing between them and the cylinder walls. 

Pumps are provided for raising water or fluids and the 
work they do in foot-pounds is equal to the weight raised 
by the height in feet through which it is raised; they 
form only a more convenient method for doing the work. 

If, for example, it be desired to lift 500 gallons of water 
at 8j lbs. per gallon from a cistern or tank to an elevated 
tank 27 feet above the cistern, the work to be done would 
be 500X8JX27 ft. = 112,500 ft.-lbs. of work, and could 
be performed with a bucket alternately filled at the 
cistern and emptied into the elevated tank; the number 
of trips from the one to the other would be determined 
by the capacity of the bucket. 

The plunger of the lift pump is no other than the 
bucket fitted to a tube and having a bottom hole provided 
with a check valve to facilitate filling. The bucket still 
must travel the full height for each delivery, and the 
tube, of course, extend the full height, also from below the 
water surface of the cistern to a point above the tank. 

It is apparent that before the bucket starts upward on 
its first trip the water level will be the same inside and 



76 Pumps 

outside the tube, likewise the pressure (atmospheric air 
14.7 lbs.) inside and outside will be alike and the bucket 
be flooded. Now, with any upward movement of the 
bucket it tends to create a vacuum (space devoid of air) 
and to destroy the atmospheric pressure (14.7 lbs.) above 
the water surface inside the tube, while the pressure on 
the water surface outside the tube has not been affected. 
Due to the unbalanced air pressure,' the water will con- 
tinue to follow the bucket upward until the pressure of 
the water column balances the difference in pressure 
inside and outside the tube. 

As shown, however, in a previous chapter the atmos- 
pheric pressure (14.7 lbs.) should support a column of 
water 34 feet — in practice, no more than 25 feet to 28 
feet is attainable due to leaks and other causes. 

If, now, the lower end of the tube be arranged with a 
suction valve, so the water could continue to follow the 
bucket or piston upward but not return, the simple lift 
pump would be complete, only the stroke would be 
excessive and impractical; by cutting off 20' to 25' of the 
lower end of tube (pump cylinder) and substituting a 
suction pipe, a short stroke only of the piston or plunger 
would be necessary. 

In the force pump with the first downward stroke of 
the solid piston the air is forced out through the discharge 
valve and passage, and the following upstroke tends to 
create a vacuum which permits the pump cylinder to fill 
with water, which is forced out through the discharge on 
next down stroke. The force pump and its arrangement 
of discharge valve, ki, permits of piping being connected to 
the discharge outlet and the water to be discharged under 
a considerably greater pressure than that due to the 
elevation of the pump; for, unlike the lift pump, which is 



Pumps 77 

limited to approximately 25' to 28' lift, its limit of pres- 
sure on the discharge is governed by the physical strength 
of the pump and the available power behind it. 

The lift pump is used principally for pumping out 
cisterns, tanks and ditches. 

The force pump is used for any purpose where the 
water or fluid is to be distributed through pipes under 
pressure, such as boiler feeding, water supply and the like. 

The Centrifugal pump, unlike the (reciprocating) lift 
and force pumps considered above, has no valves and 
no plungers, but has in place of the plunger an impeller 
wheel, driven at high speed and having vanes or spokes 
which impart centrifugal force to the water passing 
through the impeller. The water enters the pump casing 
at the center of the shaft and leaves it at the circum- 
ference. 

Centrifugal pumps must either be below the water 
surface or be primed (filled) before starting, as they 
would only churn the air and not establish the difference 
in pressure between the inside and outside of the suction 
pipe. 

They are in many cases supplanting reciprocating 
pumps due to their simplicity of construction, their 
simplicity of operation and the small amount of atten- 
tion required. 



CHAPTER V 



STRENGTH OF MATERIALS 

In the design and construction of machines or the 
several parts or elements of which they are composed 
it is necessary— not only to apply the principles of me- 
chanics as to the forces involved and their action but to 
also select the proper material and its size and shape to 
safely transmit such forces. It is evident, therefore, that 
the strength and nature of the materials to be used and 
their adaptability should be known. 

A force, depending upon its nature, acting on any one 
element of a machine may tend: (See Fig. 1.) 

To pull such member apart by means of a tensile strain, 

To crush it under a compressive strain, or 

To cut it in two under a shearing strain. 

The external or applied force tends to change the shape 
or to disrupt the part and subjects it to a "strain," as 
stated above; the force acting within the body to resist 
disruption is called "stress." 

There are, therefore, three kinds of simple or direct 
stress: 

Tensile — to resist pulling apart, as in ropes and tie-rods. 

Compressive — to resist crushing, as in columns and sup- 
ports. 

Shearing — to resist being cut in two, as in cutting or 
punching materials or plates. 

A tensile strain causes elongation. 

A compressive strain, pushing together or compression. 

A shearing strain, cutting across to separate the par- 
ticles of the material. 



80 



Strength of Materials 



Any applied force tends to deform the body or member 
and any one or all three kinds of stress may act to resist 
such deformation. 



*|Sf| 



>v 



% 



I"- 






w 



*» 

t 

o 



Z0&77Z& 



Comnrav:^ 



JZL 



<6 



Seam. 



Tats/on. 

Span — . 



FlG.i. 




UNIT STRESS 

When a body is subjected to a strain, the total stress 
inside the body is considered as distributed evenly over 
the entire cross-sectional area. The total stress divided 
by the said area in square inches or square feet is called 
the unit stress per square inch or per square foot, accord- 
ing to the terms taken for the cross-sectional area. 

If A equals the total area, P — the total stress in pounds 
and S — the unit stress; then 



A- s , 



P = AS, 



b ~A 



The unit stress for metals and timber is considered 
usually in pounds per quare inch, while for brick and 
stone and concrete it is taken in pounds per square foot. 

Any material used in construction has a certain 
rigidity due to the attraction of its particles for each 



Strength of Materials 81 

other; this mutual attraction holds the material to its 
shape and determines its strength against deformation 
when strained. The various materials used in construc- 
tion vary in strength and suitability due to differences in 
their nature — some being suitable for use under compres- 
sion but unsuitable under tension. 

HEAT 

Heat affects the strength of metals, as it tends to soften 
them and to destroy the strong attraction between the 
particles. This may be observed by taking an iron bar 
which, when cold, offers considerable resistance to bend- 
ing but upon being heated bends under its own weight. 
Metals of the same kind vary in strength, due to dif- 
ferences and imperfections in manufacture; timbers, 
etc., to their growth and treatment. Therefore, average 
values for strength of a material are usually given in the 
tables of hand books, etc. 

ULTIMATE STRENGTH 

When a materialis subjected to a gradually increasing 
strain or load and consequent stress until it is ruptured, 
the intensity of such strain in pounds divided by the area 
of the cross-section in square inches is called the ultimate 
strength per square inch, and may be tensile, compres- 
sive or shearing, depending how the load was applied. 
The ultimate strength may be, and probably is, reached 
just before rupture takes place. 

In the design of structures the ultimate strength or 
resistance against rupture should not be taken as the 
stress to which the material may be subjected. In prac- 
tice the allowable stress is taken as about 30 to 50 per 
cent, of the elastic limit of the material or to a fractional 



82 Strength of Materials 

part of the ultimate strength, such fractional part being 
called factor of safety. 

ELASTIC LIMIT 

All materials are more or less elastic and will return 
to their original shape after being deformed by the action 
of a load, provided such load has not been great enough 
to produce a permanent set or deformation. The maxi- 
mum load per unit of area to which a material may be 
subjected and still return to its original shape — -i. e., no 
permanent set has taken place — is known as the elastic 
limit of the material. 

FACTOR OF SAFETY 

Often in place of using the elastic limit for purposes of 
computation a fraction of the ultimate strength of the 
material varying from one-third to one-twentieth of such 
ultimate strength is taken; depending on the condition 
of applying the load and the nature of the material. 

MACHINES AND STRUCTURES 

Are subjected to various conditions of loading: Steady, 
intermittent, shocks, etc.; judgment, therefore, must be 
exercised in assuming the working stress or factor of 
safety to be used in any special case. 

DEFORMATION 

All materials are more or less deformed when subjected 
to a load, but immediately assume their original form, 
provided the stress produced was well within the elastic 
limit of the material. Under this condition the deforma- 
tion is practically proportional to the stress and also 
proportional to the length of the bar. 



Strength of Materials 83 

COEFFICIENT OR MODULUS OF ELASTICITY 

The ratio between the stress per unit of area to the unit 
deformation when such stress is within the elastic limit. 

If E be the coefficient of elasticity, 5 the unit stress 
and d the unit deformation, then within the elastic limit, 

E = ~, and S = EXd 

a 

If the elastic limit is exceeded, a permanent set or 
deformation results and the material does not return to 
its original shape or form. 

WORKING STRESS 

The allowable stress to which a material may be sub- 
jected without injury. It should be more than the 
weight of the member and the extraneous load applied 
thereto. A bar if sufficiently long would bend or break, 
due to its own weight, without any extraneous load. 

TENSILE STRENGTH 

The ability of a material to resist a tensile strain, or to 
being pulled apart, by an equal tensile stress. If such 
stress is well within the elastic limit, no injury to the 
material ensues and no permanent set takes place. 

Any one of a number of materials might be used to 
support a weight suspended thereby, but each would 
have to be of a cross-sectional area commensurate with 
its strength per square inch (unit area). 



84 



Tension and Compression 



STRENGTH OF MATERIALS— TENSILE 





Average in lbs. per Sq. In. 


Coefficient of 




Ultimate 


Elastic Limit 


Elasticity 


Timber 
Cast Iron. . . 
Wrought Iron 
Steel 


10,000 
20,000 
48,000 
60,000 


3,000 

6,000 

24,000 

30,000 


1,500,000 
15,000,000 
25,000,000 
29,000,000 



A working stress about .5 (elas. limit) is used, if 
the load is steady and slowly applied. If the material is 
subject to sudden loading or shocks, one-half or one-third 
of the above should not be exceeded. 

It will be seen with the above working stresses, a bar of 
cast iron of one square inch area is only one-fourth as 
strong as a similar bar of wrought iron, but approximately 
twice as strong as a similar bar or rod of wood. 
If Z = the weight in pounds to be supported, 
<z = the cross-sectional area of the bar in sq. in., and 
S = allowable working stress in pounds per sq. in., 



Then <z = ^, 



S = 



L = aXS 



L 

S' a- 

Example. What weight (steady) will a round steel bar 
2" in diameter support safely? 

Solution. L = aXS. The area equals 2" X2"X. 7854 = 
3.1416 sq. in. S = 16,000 lbs. L = 3. 1416X16,000 
or 50,266 lbs. 
Example. A weight of 48,000 lbs. is to be suspended 
by a square wrought iron bar. What size of bar is neces- 
sary if the load is steady? 



Tension and Compression 85 

Solution. a = -^, L = 48,000 lbs., S = 12,000 lbs. 

Then a = 48,000 -M 2,000 = 4 sq. in. or 2 M X2" in size. 

If a round bar had been called for, the diameter of a 
circle whose area is 4 sq. in. would have to be found. 

Example. To a suspended flat steel bar 2"X1.5 n a 
weight of 24,000 lbs. is suddenly applied. What is the 
unit stress to which the material is subjected and is the 
bar safe? 

Solution. S = -, £ = 24,000 lbs. * = 2 f! X1.5- !! = 3 sq. in. 

a 

Then S = 24,000-v-3 = 8,000 lbs. unit stress to which the 
bar is subjected. Under this loading one-half the working 
stress of 16,000 lbs. is taken, and as this allows 8,000 lbs. 
per sq. in. the bar is safe. 

In round tension rods and bolts where the load is 
applied through the medium of screw threads the cross- 
sectional area of such bars is taken at the root of the 
thread, as this area determines the strength of the bar 
or bolt. In long tie-rods the ends are usually upset to 
provide a sufficiently larger outside diameter, so that the 
diameter at the root of the thread and the diameter of 
the bar between the upset ends are about equal. 

In bolts the height of the nut is usually equal to the 
outside diameter of the threaded portion and the height 
of the head somewhat more than two-thirds the diameter. 
These proportions are used in order to have all parts 
equally strong approximately. 

Under the heading "Factor of Safety" the proper value 
of such factor for any special case was not considered; 
such, however, may be found in the accompanying table 
taken from "Machinery's" hand book. 



86 



Tension and Compression 



FACTORS OF SAFETY 







Load Varying from 


Sudden 




Steady 


to Max. 


Varying 




Load 


One 

Direction 


Both 
Directions 


Loads and 
Shocks 


Wood 


8 


10 


15 


20 


Cast Iron 


6 


10 


15 


20 


Wrought Iron . . . 


4 


6 


8 


12 


Steel . ... 


5 


6 


8 


12 



In the table the factors are given as whole numbers. 
It is, however, to be understood that the ultimate 
strength of the material is to be divided by the proper 
factor to find the allowable working stress; after which 
the method of solution is the same as if the allowable 
working stress based on the elastic limit had been taken 
directly from a table of strengths of materials. 

ROPES AND CHAINS 

These are used in tension; the student is referred to 
any of the various engineers' pocketbooks, which have 
tables giving the strength of different kinds of ropes, 
chains, cables, etc. 

COMPRESSIVE STRENGTH 

As compression is the opposite of tension, the compres- 
sive strength represents the ability of a material to resist 
the crushing action of a load acting thereon. When the 
length of a piece is short compared with its cross-sectional 
area, the material is considered as in direct compression. 
When the length is ten or more times the smallest cross 



Shear and Torsion 



87 



dimension, it is liable to fail, due to bending. Materials 
when in direct compression have approximately the same 
characteristics of the elastic limit and coefficient of 
elasticity as in tension. 

The average ultimate compression strengths, however, 
are taken as follows: 

Timber (lengthwise) 8,000 lbs. per sq. inch 

Cast Iron 80,000 

Wrought Iron. . 55,000 

Steel (castings) 100,000 

Brick 6,000 to 10,000 

Stone 10,000 to 12,000 

The formulas for direct compression are the same as 
those for tension. 



a 



a 



a 



<< 



SHEARING STRENGTH 

The ability of a material to resist being cut in two 
when acted upon from opposite sides by two members of 
a machine or structure tending to slide upon each other, 
as in the blades of scissors, shears, punches and dies, 
plates in tension, as in riveted joints, the stripping of the 
threads between a nut and its bolt, etc. 

The formula for direct shear are the same as for tension 
and compression. L, representing the weight; a, the 
cross-sectional area in shear; and S, the allowable shear- 
ing stress of the material. 

In riveted joints, where rivets prevent the sliding of the 
one plate upon the other, the area of cross-section of the 
rivet times the allowable shearing stress gives the shear- 
ing value of the rivet. In punching or in the shearing of 
plates the area to be sheared is equal to the length of 
the sheared portion multiplied by the thickness of the 
plate, both being in inches. 



a a a 

a a ct a 

a a a u 

a a a a 



88 Shear and Torsion 

Some materials, such as wood and similar fibrous 
materials, offer a greater resistance to shear when cutting 
across the grain than when cutting parallel with it. 

The average ultimate shearing strengths of various 
materials are taken as follows: 

Timber, parallel with the grain 600 lbs. per sq. in. 

Timber, across the grain. .... 2,500 

Cast Iron 18,000 to 20,000 

Wrought Iron 40,000 

Steel 70,000 to 75,000 

TORSION 

Any action which tends to twist a member or a part of 
a machine puts such part under a torsional strain; this 
must be counteracted by an equal torsional stress set 
up in such part. If the unit stress is within the elastic 
limit of the material no permanent deformation results, 
but if in excess of such limit a permanent deformation 
and possible disruption of the part results. Many of 
the common tools, such as screw-drivers, drills, socket 
wrenches, keys for locks, shafts for transmitting motion 
and power, etc., in their use are subjected to torsional 
strain. 

A shaft or tool assumed as fixed 
at one end and provided with a lever 
or crank arm to be acted upon by a 
weight or a pull (see Fig. A) would 
be subjected to a torsional strain or 
twisting moment. 

If L represents the lever arm of 
the shaft in inches, W the weight or 
pull in pounds, and T.M. the twisting moment in inch- 
pounds, then T.M. = WXL. 




Shear and Torsion 89 

Any weight on the lever arm will twist the shaft more 
or less, starting with zero at the fixed end and reaching 
a maximum where the lever is attached. The amount 
of the twist increases in proportion to the length of the 
shaft, provided the elastic limit is not exceeded. The 
lever will return to its original position also after the load 
is removed if the elastic limit was not exceeded. 

The resultant stress in a shaft under torsional strain 
is in the nature of a shear at each side of a plane cutting 
across it at any point. 

The shearing stress in a round shaft varies from at 
the center to a maximum at the circumference. 

If ^r = the safe torsional moment of resistance of a 
shaft; then RT should equal T.M. or RT=T.M.; but 
RT must equal the permissible working stress «S of the 
material in pounds per square inch multiplied by Z t , the 
section modulus for torsion; a constant determined by 
the shape of the cross-section of the shaft. 

Now, as RT = SxZ t so also must 
WXL = SXZ t 

The permissible unit working stress S may be taken 
as 9,000 lbs. per square inch for steel. 

The section modulus for torsion or the constant Z t for a 

Circle = 3 -^ 3 = 0.196* 
16 

2a 3 
Square = — - =0.22 a 3 approx. 

(See Fig. B) 

Example. Required the diameter of a solid shaft — 
provided with a lever arm 15" long, from the end of which 



90 



Shear and Torsion 



* — d 





is suspended a weight of 5,000 lbs. The 
shaft to be of steel? 

Then T.M. = WXL or 15X5,000 = 75,000 
inch-lbs., and as RT must equal 75,000 
inch-lbs. 

then also SXZ t = 75,000. 

Now, by substitution 'we have 

9,000 X0.196J 3 = 75,000 

1,764.000^ 3 = 75,000 

75,000 



and d 3 



42.517 



FJ&.B, 



1,764 

d= ^42.517=3.5" approx. 

For further study, see standard hand and 
pocket books already referred to. 



PIPES AND HOLLOW CYLINDERS 

These are used for storing or conveying fluids for 
many varied purposes. The materials used most for 
this purpose are cast iron, wrought iron and steel. In 
any case the one best suited for the purpose is selected. 
Small pipes for conveying water, gas, air, etc., generally 
are of wrought iron or steel. 

The larger pipes for gas and water and the cylinders 
of engines and pumps are usually of cast iron. 

Tanks usually are built up from wrought iron or steel 
plates riveted together at the joints. 

For steam the pipe lengths are long, narrow plates or 
sheets of wrought iron or steel rolled into a tubular form, 
after which the two edges are welded together. 



Pipes and Cylinders 



91 



It will be readily understood that the thickness of pipes 
and cylinders must be such that they will withstand the 
internal pressure to which they are to be subjected. Cast- 
iron, wrought-iron and steel pipes are made up by manu- 
facturers in standard diameters and of standard thick- 
nesses of metal. The walls of cast-iron pipe are usually 
made thicker than actually required to withstand the 
normal pressure, in order to withstand shocks in handling 
and water hammer after being placed in service. Both 
call for large factors of safety and consequent low allow- 
able working stresses. 

It has been previously shown that fluid pressure acts 
in all directions and tends to tear the containing vessel 
apart. The pressure is usually expressed in pounds per 
square inch for each square inch of surface. In a pipe 
filled with water subjected to pressure the greatest strain 
tending to tear the pipe along the side lengthwise for any 
one inch equals the product of the diameter in inches 
times one inch length times the water pressure in pounds 
per square inch. As 
water is practically 
incompressible, it will 
be readily understood 
that for any one inch 
length of the pipe 
(see Fig. 2) the water 
contained on opposite 
sides of any one diameter is subjected to a spreading 
action equal to the pressure per square inch on each inch 
of said diameter. The spreading action is resisted by the 
two thicknesses of metal, one at each end of the diameter 
and therefore each side of the pipe resists one-half the 
strain and is consequently stressed to that extent. 




92 Pipes and Cylinders 

Hence the formula, Pressure (P) X diameter (d) X 
length (/) =2 X thickness (0 X length (/)X ultimate tensile 
strength of the material (S)-f-factor of safety (/"), or 

Pdl=^- and Pd = ~ 

Example. What thickness of cast iron is required for 
a water pipe 10" in diameter under a pressure of 100 lbs. 
per square inch, with a factor of safety of 16? 

Solution. By substituting in the above equation 

1 00X10 = 2X ^f'° 00 
16 

or 1,000 = 2,500X*, and t = 0A" 
If in the above case the ends of the pipe were pro- 
vided with flanges, to which flanged covers are to be 
bolted, the pressure would tend to burst the covers and 
to tear the pipe apart circumferentially. In order to 
ascertain whether the thickness 0.4" as obtained is suffi- 
cient, also under this condition it would be advisable to 
find what factor of safety this thickness would represent. 
If this is found to be equal to or greater than the factor 
(16) used above, the walls would be considered sufficiently 
strong. 

This can readily be found by dividing the product o 
the net area of metal Xultimate strength (S) by the 
total pressure exerted against the cover. 

Formula. Pressure per sq. in. (P) X inside diam. 2 (d 2 ) X 
.7854 = [outside diam. 2 (Z> 2 )- inside diam. 2 (^ 2 )]X.7854X 
ultimate tensile strength (S)-^ factor of safety (/), or 

PX^X.7854 = - 7854( ^ )XS 

By substitution, then 100X10X 10 X. 7854 = 

.7854(1 16.64- 100) X20,000 7854X16.64X20,000 

/ and/- 10) 000X.7854 

= 33.28 



Riveted Joints 93 

The factor as found is more than twice that assumed 
as necessary for the walls of the pipe; the walls, therefore, 
would be considered as of ample thickness. 

The difference in area between the outer diameter 
circle and the area of the inner diameter circle represents 
the area of metal for resisting the strain. 

The formula for all practical purposes, if the walls are 
relatively thin as compared to the diameter, may be 
expressed in the simple form: 

pw 4X/XS 
rXd = 7 or 

4X0.4X20,000 
/_ 1000 ~ d u 

for the example as given above. 

This formula holds good also for thin hollow spheres, 
the inside diameter being taken as the counterpart of the 
inside diameter of the pipe. 

RIVETED JOINTS 

Rivets when used to join two plates which tend to 
slide one upon the other or to be pulled apart length- 
wise are subjected to compression, tending to crush 
them where they pass through the individual plates and 
to shear or cutting across at the point where they pass 
from the one plate into the other. 

Bearing Value of Rivets. — The bearing value is usu- 
ally taken as equal to the rivet diameter multiplied by 
the thickness of the plate multiplied by the safe bearing 
strength of the material of the plate or rivet, depending 
which has the lower value or compressive strength. 

Rivet Shear. — The shearing strength of the rivet is 
taken as equal to the area (diam. 2 X.7854) of the rivet 



94 



Riveted Joints 




F/6. 3. 



multiplied by the safe shearing strength of the rivet 
material. 

One of three types of riveted joints is generally used 
in practice, namely: 

1st — Single riveted lap joint (see 
Fig. 3) in which the edge of one 
plate laps the edge of the other 
plate, with a single row of rivets 
fastening the two plates together. 
Here each rivet is required to 
resist the tensile strain of the plate 
for a width equal to the center to 
center distance or pitch of the 
rivets. 

If P equals the strain transmitted from the plate to 
each rivet as above, t the thickness of the plate, d the 
diameter of the rivet, p the pitch of the rivets or center 
to center distance between them, S t the tensile stress, S b 
the bearing value and S s the shearing stress, each per 
square inch, respectively, for plates and rivets for the 
narrow strip represented by the pitch of the rivet; 
then the resultant stresses produced in single riveted 
joints would be: 

Tensile Stress, (S t ) = strain (P) divided by the dif- 
ference between the products of the plate thickness (0 
times the pitch (p) and the thickness (t) times the diame- 
eter (d). Then the unit tensile stress in the plate, 
P P 

1 tp — td l t{p — d) 

Bearing, S\> = strain, divided by the thickness times 
the diameter of the rivet; equals the unit compressive 
stress of the plate and rivet, or 

s - p 



Riveted Joints 



95 




F/6.4- 



Shearing Stress, S 8 = strain, divided by the area of 
the rivet (.7854 d 2 ) or 

S - P 

° 8 .7854 </ 2 

2nd — Single Shear double riveted joints, in which two 
rows of rivets, staggered are used. (See Fig. 4) 

The stresses in double riveted 
joints (single shear) are as follows: 

Tensile Stress, this is the 
same as for single riveted joints. 

Bearing, S^ = strain, divided by 
twice the thickness of the plate 
times the diameter of the rivet, or 

s - p 

ShearingStress,S s = 2x78 P 54x ^ 

3rd — Butt joints in double shear, in which the edges 
of the plates butt one against the other instead of lapping 
one upon the other (See Fig. 5). 
A joint of this kind has two 
cover or splice plates, one at 
each side and each of which 
laps over the abutting plates 
or plates to be joined. The 
rivets at each side of the joint 
pass through the two cover 
plates and through the plate to 
be joined, lying between them; 
that is, they pass through the three thicknesses of plate. 
It will be seen that the rivets are all placed in double 
shear or in shear between each face of the plate to be 
joined and the adjacent cover plate. 




FIG. 5 



s„= 



96 Riveted Joints 

The stresses in butt joints, single riveted, double shear; 
Tensile Stress; the same formula for single riveted 
lap joints is used or 

o- P 

4 Kp-d) 

Bearing. — The same formula as for single riveted lap 
joints in like manner holds good here, also; or 

P 
td 

In both cases the thickness of the joined plates is the 
determining factor, as the thickness of each cover plate 
ranges usually from 75 to 100 per cent, of the plates to be 
joined. 

Shearing Stress. — The formula for the shearing stress 
of the rivets is, however, the same as that for a double 
riveted lap joint, as the plates tend to cut through the 
rivets at two separate points. Each rivet in double shear 
acting the same as two rivets in single shear, and 

S - P 

s 2X.7854X^ 2 

In the above joint it is true there are two rows of 
rivets, but each row serves only one of the joined plates, 
and the joint is consequently single riveted as compared 
to a double riveted joint; the rivets are, however, in 
double shear, as against single shear in single riveted 
joints. 

Butt joints are made also as double and triple riveted 
for additional strength. In these cases one of the cover 
plates is usually wider than the other, and the outer rows 
of rivets are placed in single shear, as they pass through 
one joined plate and one cover plate only. 

All holes in riveted joints should be of full bore through- 
out and true, as offset holes reduce the shear section of 
the rivet and likewise weaken the joint. 



Riveted Joints 



97 



For further information on riveting and riveted joints 
the student is referred to the various engineers' hand 
books and steel manufacturers' pocketbooks. 



Allowable stresses for some materials, as taken from 
the Building Code of Baltimore, Md., issue of 1908, are 
given in the following table: 

Maximum Allowable Stresses for Various Building Materials. 
Baltimore City Building Code— 1908 



Material 


Unit Stress or Pounds per Square Inch. 


Tension 


Com- 
pression 


Shear 


Bending 

Ext. 

Fibres 


Bearing 


Rolled Steel 


16,000 
16,000 
12,000 

5,000 


16,000 
16,000 
12,000 

16,000 


9,000 


16,000 




Cast Steel 




Wrought Iron 

Cast Iron (Short 
Blocks) 


6,000 

3,000 

10,000Shop 
8,000Field 
7,000Field 

7,500Shop 

6,000Field 
5,500Field 

100 

500 

85 
350 
100 
720 

90 
400 










Steel — Pins & Rivets. 




20,000 


Steel — Pins & Rivets. 






Steel Bolts 










Wrt. Iron — Pins & 
Rivets 






»•»•••• 


15,000 


Wrt. Iron — Pins & 
Rivets 








Wrt. Iron Bolts 










Long Leaf Pine (with 
Grain) 


1,800 


1,000 

600 
800 
400 
1,000 
600 
800 
400 


1,800 




Long Leaf Pine 

(across Grain) 




White Pine (W. Gr.) . 
White Pine (A. Gr.).. 


1,000 


1,000 




Oak (with Grain) .... 
Oak (across Grain).. . 


1,500 


1,500 





Virginia Pine(W.Gr.) 
Virginia Pine (A.Gr.) 


1,200 


1,350 











98 Beams 



BEAMS 



A long piece of timber or metal (steel, iron, etc.) sup- 
ported at one or more points for carrying loads or 
weights and transferring the action resulting therefrom 
to the beam supports, the loads being off to one side and 
not immediately above or in line with the beam supports. 

Examples. The horizontal timbers. or joists and beams 
for supporting the floors of buildings, weighing scale 
beams, the walking beam of side wheel steamers, levers 
in machines, etc. 





FI6.6. FI6.7. A ' FIG. 8. 



General Classes: (See Figs. 6-7-8) 

A Simple Beam. — A beam fixed or supported at both 
ends to span the space between such supports. 

A Cantilever Beam. — A beam having one end fixed 
in a support and the other end free. 

A Continuous Beam. — A beam extending over two 
supports and some distance beyond each with the said dis- 
tances equal. These, however, will not be considered 
here, as their study requires more advanced knowledge. 

Reaction of the Support. — A stress exerted by the 
support to balance the combined force due to the weight 
of the beam and the load carried thereby. The weight of 
the beam is usually small as compared to load or weight 
supported thereby and is frequently neglected. It is 
plain the support must be strong enough to resist the 
force acting thereon, otherwise the structure would be 
destroyed. The total stress of the support or supports 
is equal to the load. If there is more than one support, 



Beams 99 

the nature and relative location of the load, on the basis 
of the theory of levers, determines what part of the load 
is carried by each support. In the case of the cantilever 
it is plain that the one support carries the entire load, 
including the weight of the beam. There is, however, a 
tendency also to topple over the support and bend the 
beam anywhere throughout its length, especially at the 
support. The extent of this tendency at any point of 
the beam is called the Bending Moment at such point. 

In the case of Cantilevers. If W=th.e weight, l=the 
distance from the support in inches, and ^ = the sup- 
port, and F r = the reaction of the support; then F r =W. 

In the case of the Simple Beam. If W= weight, 1 = dis- 
tance from the center of the weight to the left-hand 
support; li = the distance to the right-hand support; 
A and A\ the left hand and right hand supports, respec- 
tively, and likewise F, Fi the force exerted on each one; 
then by considering the beam as a lever the pressure or 
force F, Fi on either support may readily be found. 

In considering the beam as a lever, first one support 
A, or Ai, is taken as the fulcrum and then the other; the 
weight arm being the distance from the weight to the 
fulcrum and the power arm the distance between the two 
supports. This may be expressed by the following 
formula: 

~ , £ 4 N Weight (/F)Xdistance (1) 

Mf0rCe ° Ul) = Length of the lever (1 + U) ° r 

v WXl 



1 + 

And.F = 



rxii 



1 + 1: 



100 Beams 

Example. Let the weight (W) equal 500 lbs. The dis- 
tance 1, from one support to the weight equal 1 foot 
(distance expressed in inches). The distance li from 
the weight to the other support equal 4 feet (expressed in 
inches); then the reaction of support A\ equals 

r 500X1'X12" innl , . 

^ 1 = — civnii — = I™ lbs. reaction A\ 

. , „ 500X4X12 M 4AA1 , . . 

And / = — t ,, — =400 lbs. reaction A 

o X 1Z 

As the sum of the two reactions can be neither more 
nor less than the total load, then A-\-A\ must equal 500 
lbs., and is found to do so. 

The distances from the weight or weights to the ful- 
crums, when finding the reactions only, may be expressed 
in feet, but in view of other formula it is considered more 
advisable to take them in inches also. In either case all 
measurements must be taken in the same unit (feet or 
inches). 

In machines, beams in the form of levers are used 
extensively both as cantilevers and plain beams. The 
tooth of a gear outside the rim, also the crank of an 
engine, are cantilevers. The iceman's scales, the common 
seesaw, are, in fact, two cantilevers fixed rigidly together 
but extending from opposite sides of the support or 
fulcrum. A shaft extending from one hanger to another 
but not beyond, on which are mounted pulleys or gears, 
is an example of the simple beam. The pulleys tend to 
bend the shaft, due to their weight and the pull of the 
belt thereon. 

If, in the example of the 500-pound load on the 5-foot 
length beam, the point of application of the load had 
been taken as the fulcrum of a lever and under a force 



Beams 101 

of 500 pounds tending to move it, but held against 
movement by the action of weights or forces at the ends 
of the two lever arms, it is evident first that the one 
weight would have to balance the other and that the 
combined weight of the two would have to equal the 
force of 500 pounds tending to move the lever and weights 
bodily from their position and all in the same direction. 

In any simple beam the greatest tendency for it to 
bend is at the point of greatest or maximum load. In 
the case referred to this would be at the point of applica- 
tion of the weight, whether this point had been considered 
as the fulcrum or as the point of application of the load. 

The force tending to bend a beam at any point is known 
as the bending moment at such point, and the maximum 
bending moment is consequently at the point of applica- 
tion of the maximum load. 

In cantilevers the fulcrum is at the point of support, 
and the action of the forces tending to bend the beam 
is, therefore, greatest at this point and is the point of 
maximum bending moment of the beam or cantilever. 
This maximum bending moment in cantilever may be 
expressed by a formula, by applying the principle of the 
lever, as follows: 

Let B.M. (max.) represent the maximum bending 
moment at the point of support; W (the weight in 
pounds); and 1 (the distance from the support to the 
point of application of the weight, in inches); then B. M. 
(max.) = /FXl, in inch-pounds. 

Bending Moment. — The force acting at any point of 
a beam tending to rupture it by tension, pulling apart 
at the bottom side usually, and by compression, pushing 
together at the top or opposite to the side in tension. 
Between the two somewhere and parallel therewith there 



102 Beams 

is a plane in the beam where the fibers are neither in 
tension nor compression. This plane passes through a 
point in the cross-section of the beam called the neutral 
axis. 

Shear. — Beams, too, are subjected to shearing strains, 
due to their weight and the action of the loads, which 
tend to cut them in two crosswise, and is, of course, 
greatest at the points of support; this shearing tendency 
must be resisted and balanced by the shearing stress of 
the beam, while the supports must likewise counteract 
the crushing action to a like amount. 

Loading of Beams. — In the foregoing one method of 
loading has been considered. The load may consist of a 
single weight concentrated at one point; it may be 
evenly distributed over the full length of the beam or it 
may consist of a number of loads located at various 
distances from the supports. The method of loading 
materially affects the intensity of the bending moment, 
and therefore each problem must be considered under its 
special class. 

Formulas have been evolved for finding the maximum 
bending moments and shears for various kinds of loading, 
some of which are given below — in which B.M. is the 
bending moment; 1 the length of the span in inches; W 
the weight in pounds; F s the shear; and A the support. 

It should be remembered that the weight of the beam 
acts also as a part of the total load, and is taken as a dis- 
tributed load equal to the weight of the beam, dis- 
tributed over its entire length. Only the extraneous or 
outside load is here considered for the sake of clearness. 
Each load and its action are computed separately and 
the results added together to obtain their combined 
action. 



Beams 103 

Simple Loading Conditions of Beams 

1st. Cantilever beam, one end fixed, with a single load 
at the opposite or free end. 

Max. B.M. =WX1. At support. 



^ A r/e.s. 6" Max. F s = W. At support. 

2nd. Cantilever beam, one end fixed, with a load 
evenly distributed over the full length of the beam. 

^^^^w^vv , Max. B.M. =/FXjl. At support. 

Max. F s = W. At support. 



I- 



FIG. /&. 



3rd. Simple beam, fixed at both ends and loaded at 
the center. 



Max. B.M. =iWX 1. At load. 
Max. F s —\W. At supports. 



*—*- — f 



FI6. If. 



4th. Simple beam, fixed at both ends, with a load 
evenly distributed over the full length of the beam. 

^sssass^^^ ^^^ Max. B.M. =J/FX/. At load. 
( lj~ Fie.js v Max. F s —\W. At supports. 

After the maximum bending moment is obtained it 
becomes necessary to select a beam whose safe moment 
of resistance R will equal or be greater than said bending 
moment, and whose shear resistance is greater than the 
shear produced by the load. The latter may be found by 
dividing the shear by the safe unit value allowed for 
shear to ascertain the required cross-sectional area 
necessary to resist such shear and comparing it with 
that of the beam whose safe moment of resistance has 
been found sufficient to resist the maximum bending 
moment. 



104 



Beams 



Moment of Resistance. — The moment of resistance, 
R, may be determined by multiplying two factors to- 
gether; one, the safe stress, /, of the material, and the 
other a constant, S, called the modulus of the section. 
This constant depends upon the form and dimensions of 
the cross-section of the beam. 

The moment of resistance, R=fXS 



Section Modulus for Simple Sections or Shapes 

A Rectangle, where d = height in inches, 

b — width in inches, 
and S = section modulus; 
bXd 2 




AkiS 



then S = 



A Circle 



S = 



3.1416X^ 3 



32 



FIG. /4. 



Axis 




A Hollow Rectangle S = 



bXd*-hXd\ 
6Xd 



A Hollow Circle 



S = 



3.1416X(W 4 -^) 



32Xd 



Example. A plain rectangular bar of steel is to support 
two loads, one of 600 lbs. concentrated at the center of 
the span (Case 3), and the other of 1,200 lbs. evenly dis- 



Beams 105 

tributed over the bar (Case 4). The span is 4 feet. What 

size of bar will be required to support the weight safely 

if 16,000 lbs. be allowed for the stress in tension, and 

16,000 lbs. for compression and 10,000 lbs. for shear? 

Solution. B.M. (max.) for the 600 lbs. weight equals 

600X48 in. . 
= 7,200 inch-pounds, 

B.M. (max.) for the 1,200 lbs. weight equals 
1,200X48 in. 



7,200 inch-pounds. 



8 

Combined max. bend, moment = 7,200 + 7,200= 14,400 
inch-pounds. 

Now, as the moment of resistance (R) must equal the 
max. B.M. and R=fXS, then also must /X<S = max. 
B.M., but the B.M. in this particular case is 14,400 lbs. 
Therefore, 14,400=/XS; but /= 16,000, then 

14 400 
14,400 = 16,000X5, and S = — , —— or 0.9 

16,000 

b X d 2 
Now S = — - — , and by substitution and clearing the 

fraction £X^ 2 = 0.9X6 or 5.4. 

If, now, the thickness is assumed as % in. or .625", 
then .625 ^ 2 = 5.4, and d 2 = 5.4-=-. 625 or 8.64, and d = 
\/8.64 or approximately 2.94 inches. The nearest stand- 
ard bar would be ^ n X3" or ^g" thick by 3 U wide. 



The shear at each support is equal to § the total load 
of 600+1,200 or 1,800 lbs. -5-2 = 900 lbs. Which is less 
than 1-10, the allowable stress in shear. 

The usual practice is to use rolled shapes, such as 
/-beams, as they require, due to their shape and design, 
less metal and have greater stability relatively than 
plates or bars on edge for two reasons, viz: 



106 Beams 

1st. The greater portion or bulk of the metal is placed 
in the flanges and furthest away from the neutral axis, 
or where it will add most to the strength of the beam. 
The beam is made high, relative to the width of the 
flanges, for the same reason that a thin board on edge is 
stiffer and stronger than when placed or laid flat. 

2nd. The flanges, even though they are much narrower 
than the height of the beam, are approximately eight 
times the thickness of the web and, therefore, afford a 
good bearing. 

In the case of the bar for supporting the 600 and 1,200 
pound loads already considered the ^ n X3" bar as found 
would safely sustain the load placed upon it, but might 
deflect or sag too much. The beams and joists in build- 
ings for supporting floors and plastered ceilings should 
be stiff and should not sag or spring but a small fraction 
of an inch when loaded, as the sagging, springing or vibra- 
tion would cause the plastered ceiling to crack. 

Formula for determining the deflection (D) of beams of 
various shapes may be found in the various hand books 
published by the large steel manufacturers. They take 
into account the loads, their method of application and 
certain properties of the shape of the beam called the 
moment of inertia, and also the coefficient or modulus of 
elasticity of the material of the beam. 

The formula for deflection (D) of a simple beam evenly 
loaded and rectangular in cross-section and used on edge 
may be expressed as follows: if ^T = the weight in pounds; 
1 the span in inches; E the coef. of elasticity, which for 
steel is taken as 29,000,000; / the moment of inertia, 
based on the kind of cross-section of the beam; b the 
thickness of the beam in inches; and the figures 5, 384, 
and 12 as numerical constants. 



D (Max. Defl. in In.) 
in which / = 



Columns 107 

5X/FX1 3 



384X£X/' 
bXd* 



12 

The value of / (moment of inertia) is found first and 
then substituted in the other formula in the place of /. 

The student may wonder why one formula is not made 
to cover all cases and thereby simplify the work. This is 
hardly possible when the many forms and combinations 
of loading and shapes of sections are considered. 

The hand books already referred to are for the purpose 
of assisting the designer and to facilitate his work. In the 
books not only are the various formulas given but also 
tables showing the values or properties of the many 
shapes and tables showing the carrying capacity of beams 
of different materials and for different lengths of span. 

The student is earnestly advised to familiarize himself 
with the use of such pocketbooks and the data contained 
therein. Limited space has permitted of explaining only 
the meaning and application and the reason for the vari- 
ous terms and formulas as given in the hand books. 

COLUMNS OR STRUTS 

In a machine or structure such parts or members acting 
endwise to transfer a direct compression load from one 
point to another, such as the columns for supporting the 
floor beams and joists and the roof trusses of buildings, 
the piston and connecting rods of engines, the supports 
or legs of a machine, a table, etc. Columns and struts 
are the opposite to ties, which act in tension to prevent 
pulling apart, while columns and struts act in compres- 
sion against crushing. In some machines, as in steam 
engines, the piston rod and connecting rod act alter- 



108 



Columns 




Fie. i7- 



nately in tension and compression as the piston moves 
backward and forward in its operation. 

The strength of columns depend upon the nature of 
the material; the cross-sectional area; the way in which 
the material is distributed about the neutral axis; the 
length of the column as compared to its type of cross- 
section; the strength of the material and the nature of 
the ends, which latter influence the strength of the column 
to a very great extent. 

The ends may be round (see Fig. 17), 
as shown at a, or one end may be round 
and the other flat as at b, or both ends 
may be flat as at c. The form c is most 
used as in buildings and supports for 
machines, and is the strongest. Form b 
and form a are used in compression members of bridge 
trusses and machine parts, connecting rods, etc. 

Relatively, a column with ends of type 
b is 2j times, and c is 4 times as strong 
as with type a; further, it has been found 
that three columns of the same cross-sec- 
tion, and with ends respectively as a, b 
and c, are of the same strength if their 
lengths vary as 1 to \\ and 2. Column a 
with the rounded ends is the shortest. 
(See Fig. 18.) 

Columns are made of quite a number of different kinds 
of material, but the more commonly used are timber, cast 
iron and steel. When of timber they consist usually of 
one solid piece; when of cast iron they are in one piece 
and cast hollow, except when acting as an integral part 
of the frame of machines, where the cross-section may be 
of almost any form to meet conditions; when made of 



/ b 



/; C 



FJ6. 18 



Columns 109 

steel, as in engines and machines, for connecting rods, 
piston rods, etc., they are usually solid and forged; steel 
pipe is often used for columns. In buildings and struc- 
tures, however, the columns consist either of a single- 
rolled steel shape or section or a combination of such 
shapes or sections, either riveted directly together or 
held in definite relationship to each other by web plates 
of by lattice or diagonal bracing riveted to the main 
sections. In the case of the web plates they also add con- 
siderable strength in one direction. 

Columns made up of more than a single-rolled shape 
are known as "built-up columns." The main advan- 
tages of built-up columns are: 

1st. The metal may be so placed that the greater por- 
tion will be located away from or distant from the 
neutral axis. 

2nd. To provide greater and more stable end bearings, 
and so distribute the weight over a greater surface. 

The formulas in general use for strength of columns are 
empirical, and are based on the assumption that columns 
may fail by direct compression, by bending and compres- 
sion combined, or by bending alone. The formulas agree 
practically with the results obtained by experimenting 
with columns of different kinds. 

In the hand books, to which attention has already been 
directed, will also be found formulas for finding the 
moment of inertia, radius of gyration, area, etc., called 
properties or elements, appertaining to the various shapes 
(steel) and the conditions under which the shapes are 
strained and exert stress. 

Moment of Inertia. — This is based on the area of the 
material of the section, the location of the neutral axis to 
the center of gravity of the section and the shape of the 
section. The neutral axis in built-up columns does not 



110 Columns 

necessarily pass through the center of gravity of the sec- 
tions (individual) of the column, but through the center 
of the area of the column, and in such the two axes are 
taken at right angles to each other. Columns due to their 
shape in cross-section are usually more rigid laterally in 
one direction than in a direction at right angles thereto, 
because the distance from the neutral axis to the extreme 
fibres or outside edges in one direction is greater than 
the distance from the other neutral axis to the extreme 
fibers. This is evident when using a thin board as a strut 
or column. The strength of a column, therefore, is in a 
large measure determined by its least width of cross- 
section. 

It is also evident from the above that there is a moment 
of inertia for each neutral axis. 

Radius of Gyration. — This term is used to express 
the relation between any moment of inertia and the area 
of a shape, section or column under consideration. Its 
use is to facilitate finding the safe load any shape or 
section will sustain when used as a strut or column. 

If r = radius of gyration, / = moment of inertia and A = 
area, the inch being the unit of measurement, then 

For a rectangular column section with the neutral axis 
through the center of gravity: 

b X d z 
I (moment of inertia) = , A (area.) =bXd 



, jl . IbXd* ~ 

and as r = \l-j, then r — \— 77; — '-bXd or 

= 4 uibxd or VlJ from which follows; 



r= rrx or 0.288675 Xd 



Columns 111 

Ratio of Slenderness of columns equals the unbraced 
length /, in inches between the lateral supports, divided 

by the radius of gyration r. This ratio, -, is used to find 

the allowable unit fibre stress/, for the material. 

For steel columns, the American Bridge Co. give the 

formula for allowable fibre stress, f= 19,000— 100-. It 

r 

must not exceed a maximum of 13,000. In no case 
should the ratio exceed 200. 

Example: Let / = 240" and r = 2, then /= 19,000- 

100 (^) = 7,000. 

The Baltimore City Code permits a maximum ratio 
of 120 with r, as the least radius of gyration. 



